Q: A faulty thermometer has its fixed points marked \${ 5 }^{ o }\$ and \${ 95 }^{ o }\$. This thermometer reads the temperature of a body as \${ 59 }^{ o }\$. Then the correct temperature on Celsius scale is

For two scales of temperature\$\cfrac { C-0 }{ 100-0 } =\cfrac { Reading }{ UFP-LFP } -LFP\$where LFP and UFP are lower and upper fixed points on the scale\$\therefore \cfrac { C }{ 100 } =\cfrac { 59-5 }{ 95-5 } =\cfrac { 54 }{ 90 } =0.6\$\$\Rightarrow C=0.6\times 100={ 60 }^{ o }\quad \$

Q: The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

The conversion formulae are \$t_{c}=t_{k}-273.15\$\$t_{f}=\dfrac{9}{5}t_{c}+32\$Substituting the given values, we get \$\text{Neon:}\ 24.57\ K=-248.43\ ^o C = -415.44\ ^o F\$\$\text{CO:}\ 216.55\ K=-56.45\ ^o C=-69.8\ ^oF\$

Q: The freezing point on a thermometer is marked as \$20^o\$ and the boiling point as \$150^o \$. A temperature of \$60 ^oC\$ on this thermometer will be read as

Given: Freezing point on thermometer \$=20\$ , Boiling point on thermometer \$=150\$ ,Hence, number of equal parts on thermometer \$=150-20=130\$, temperature on celsius scale \$=20^{o}C\$ Let temperature on thermometer \$=t\$ ,Therefore , \$\dfrac{t-20}{130}=\dfrac{60-0}{100}\$ ,\$\Rightarrow\$ \$t-20=78\$ \$\Rightarrow\$ \$t=78+20=98^o\$

Q: In terms of Charles' law, explain why - 273 \$\displaystyle ^{\circ}C\$ is the lowest possible temperature.

At \$0 \ K\$ or \$\displaystyle -273^oC \$, the volume of the gas becomes equal to zero . The gas ceases to exists. Hence, \$\displaystyle -273^oC\$ is the lowest possible temperature.

Q: If a thermometer reads freezing point of water as \$20^{\circ}C\$ and boiling point at \$150^{\circ}C\$ how much thermometer read when the actual temperature is \$60^{\circ}C\$.

We let the unknown degree scale be called \${x}^{\circ}\$Now we know that the freezing point\$={20}^{\circ}x\$The boiling point\$={150}^{\circ}x\$This also means that \${20}^{\circ}x= {0}^{\circ}\$ C(\$\because\$ on a Celcius Scale, the freezing point is \${0}^{\circ}\$)And \${150}^{\circ}x={100}^{\circ}\$C (On a Celcius Scale, boiling point is \${100}^{\circ}\$)Also \${(150-20)}^{\circ}x= {100}^{\circ}\$C (\$\because\$ on a Celcius Scale \${100}^{\circ}-{0}^{\circ}={100}^{\circ}\$)So, if \${100}^{\circ}\$C\$={130}^{\circ}x\$Then \${1}^{\circ}\$C\$={1.3}^{\circ}x\$To go from \$C\$ degree to \$x\$, we need to use:\$x=1.3C+20\$ And \$C=\dfrac{(x-20)}{1.3}\$So if the temperature given is \${60}^{\circ}\$C, then we calculate \$x\$ as:\$C=\dfrac{(x-20)}{1.3}\$ \$\Rightarrow 60=\dfrac{(X-20)}{1.3}\$ \$\Rightarrow 60\times 1.3=x-20\$ \$\Rightarrow 78=x-20\$ \$\Rightarrow x= {98}^{\circ}\$ So \${60}^{\circ}\$C will be equal to \${98}^{\circ}\$

Q: On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39W and 239W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39C on the Celsius scale ?

Given,\$T_{f}=39^0W\$\$T_{b}=230^0W\$We know that,\$\dfrac{T_b-T_f}{100^0C-0^0C}=\dfrac{T-39^0C}{39^0C}\$\$\dfrac{239-39}{100}=\dfrac{T-39}{39}\$\$\dfrac{200}{100}=\dfrac{T-39}{39}\$\$T-39=2\times 39=78^0W\$\$T=39+78=117^0W\$The correct option is D.

Q: At what temperature, the Fahrenheit scale reading is double of Celsius Scale reading?

Step1: Relation between Fahrenheit scale and Celsius Scale\$F=\dfrac{9}{5}C+32\$ ........(i)Step2: Applying Given condition\$2C=F\$ (given) .........(ii)\$\therefore 2C=\dfrac{9}{5}C+32\$ .......(from (i) & (ii))\$\therefore 0.2C=32\$\$\therefore C=16\times 10\$\$\therefore C=160^oC\$Hence at \$160^oC\$ Fahrenheit scale reading is double & celsius scale reading.

Q: Two absolute scales \$A\$ and \$B\$ have triple points of water defined to be \$200 A\$ and \$300 B \$ . A temperature is measured on these scales as \$T_A\$ and \$T_B\$ What is the relation between \$T_A\$ and \$T_B\$ ? Given, that triple point of water as \$273.16 K\$ .

Triple point of water on absolute scale \${\rm A}\$ and \$B\$ are \$200A\$ and \$300B \$respectively. The triple point \$A\$ in Kelvin is equal to \$273.16\$. Hence \${T_1} = {T_k}\$ \$200A = 273.16\$ Similarly, \$300B = 273.16\$ By solving above equations we get, \$\dfrac{{273.16}}{{200}} \times {T_A} = \dfrac{{273.16}}{{350}} \times {T_B}\$ Hence the ratio of the \${T_A}:{T_b} = 4:7\$

Q: The temperature of a body on Kelvin scale is found to be \$X\ K\$. When it is measured by a Fahrenheit thermometer, it is found to be \$X^o F\$. Then \$X\$ is

We know, \$F=1.8T-459.67\$Where, \$F\$=Fahrenheit Temperature\$T\$=Kelvin TemperatureNow, \$F=X°F\$ and \$T=XF\$, putting there we get,\$X=1.8X-459.67\$\$\therefore X=\cfrac { 459.67 }{ 0.8 } \\ =574.25\\\$

Question 1

At what temperature, the Celsius and Fahrenheit scales will read the same ?

Accepted Answer

The formulas for converting between degree Celsius and degree Fahrenheit are:$&#176;F = (&#176;C * 9/5) + 32$$&#176;C = (&#176;F - 32) * 5/9$To find the temperature when both are equal, we use an old algebra trick and just set &#186;F = &#186;C and solve one of the equations.$&#176;C = (&#176;C * 9/5) + 32$$&#176;C - (&#176;C * 9/5) = 32$$-4/5 * &#176;C = 32$$&#176;C = -32 * 5/4$$&#176;C = -40$$&#176;F = (&#176;F * 9/5) + 32$$&#176;F - (&#176;F * 9/5) = 32$$-4/5 * &#176;F = 32$$&#176;F = -32 * 5/4$$&#176;F = -40$So the temperature when both the Celsius and Fahrenheit scales are the same is $-40 degrees.$

Question 2

A faulty thermometer has its fixed points marked ${ 5 }^{ o }$ and ${ 95 }^{ o }$. This thermometer reads the temperature of a body as ${ 59 }^{ o }$. Then the correct temperature on Celsius scale is

Accepted Answer

For two scales of temperature$\cfrac { C-0 }{ 100-0 } =\cfrac { Reading }{ UFP-LFP } -LFP$where LFP and UFP are lower and upper fixed points on the scale$\therefore \cfrac { C }{ 100 } =\cfrac { 59-5 }{ 95-5 } =\cfrac { 54 }{ 90 } =0.6$$\Rightarrow C=0.6\times 100={ 60 }^{ o }\quad $

Question 3

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Accepted Answer

The conversion formulae are $t_{c}=t_{k}-273.15$$t_{f}=\dfrac{9}{5}t_{c}+32$Substituting the given values, we get $\text{Neon:}\ 24.57\ K=-248.43\ ^o C = -415.44\ ^o F$$\text{CO:}\ 216.55\ K=-56.45\ ^o C=-69.8\ ^oF$

Question 4

The freezing point on a thermometer is marked as $20^o$ and the boiling point as $150^o $. A temperature of $60 ^oC$ on this thermometer will be read as

Accepted Answer

Given: Freezing point on thermometer $=20$ , Boiling point on thermometer $=150$ ,Hence, number of equal parts on thermometer $=150-20=130$, temperature on celsius scale $=20^{o}C$ Let temperature on thermometer $=t$ ,Therefore ,            $\dfrac{t-20}{130}=\dfrac{60-0}{100}$   ,$\Rightarrow$ $t-20=78$ $\Rightarrow$ $t=78+20=98^o$

Question 5

In terms of Charles' law, explain why - 273 $\displaystyle ^{\circ}C$ is the lowest possible temperature.

Accepted Answer

At $0 \ K$ or $\displaystyle -273^oC $, the volume of the gas becomes equal to zero . The gas ceases to exists. Hence, $\displaystyle -273^oC$ is the lowest possible temperature.

Question 6

If a thermometer reads freezing point of water as $20^{\circ}C$ and boiling point at $150^{\circ}C$ how much thermometer read when the actual temperature is $60^{\circ}C$.

Accepted Answer

We let the unknown degree scale be called ${x}^{\circ}$Now we know that the freezing point$={20}^{\circ}x$The boiling point$={150}^{\circ}x$This also means that ${20}^{\circ}x= {0}^{\circ}$ C($\because$ on a Celcius Scale, the freezing point is ${0}^{\circ}$)And ${150}^{\circ}x={100}^{\circ}$C (On a  Celcius Scale, boiling point is ${100}^{\circ}$)Also ${(150-20)}^{\circ}x= {100}^{\circ}$C ($\because$ on a Celcius Scale ${100}^{\circ}-{0}^{\circ}={100}^{\circ}$)So, if ${100}^{\circ}$C$={130}^{\circ}x$Then ${1}^{\circ}$C$={1.3}^{\circ}x$To go from $C$ degree  to $x$, we need to use:$x=1.3C+20$  And $C=\dfrac{(x-20)}{1.3}$So if the temperature given is ${60}^{\circ}$C, then we calculate $x$ as:$C=\dfrac{(x-20)}{1.3}$  $\Rightarrow 60=\dfrac{(X-20)}{1.3}$  $\Rightarrow 60\times 1.3=x-20$  $\Rightarrow 78=x-20$  $\Rightarrow x= {98}^{\circ}$  So ${60}^{\circ}$C will be equal to ${98}^{\circ}$

Question 7

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39W and 239W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39C on the Celsius scale ?

Accepted Answer

Given,$T_{f}=39^0W$$T_{b}=230^0W$We know that,$\dfrac{T_b-T_f}{100^0C-0^0C}=\dfrac{T-39^0C}{39^0C}$$\dfrac{239-39}{100}=\dfrac{T-39}{39}$$\dfrac{200}{100}=\dfrac{T-39}{39}$$T-39=2\times 39=78^0W$$T=39+78=117^0W$The correct option is D.

Question 8

At what temperature, the Fahrenheit scale reading is double of Celsius Scale reading?

Accepted Answer

Step1: Relation between Fahrenheit scale and Celsius Scale$F=\dfrac{9}{5}C+32$ ........(i)Step2: Applying Given condition$2C=F$ (given) .........(ii)$\therefore 2C=\dfrac{9}{5}C+32$ .......(from (i) & (ii))$\therefore 0.2C=32$$\therefore C=16\times 10$$\therefore C=160^oC$Hence at $160^oC$ Fahrenheit scale reading is double & celsius scale reading.

Question 9

Two absolute scales $A$ and $B$ have triple points of water defined to be $200 A$ and $300 B $ . A temperature is measured on these scales as $T_A$ and $T_B$ What is the relation between $T_A$ and $T_B$ ? Given, that triple point of water as $273.16 K$ .

Accepted Answer

Triple point of water on absolute scale ${\rm A}$ and $B$ are $200A$ and $300B $respectively.&#10;&#10;The triple point $A$ in Kelvin is equal to $273.16$.&#10;&#10;Hence ${T_1} = {T_k}$&#10;&#10;$200A = 273.16$&#10;&#10;Similarly,&#10;&#10;$300B = 273.16$&#10;&#10;By solving above equations we get,&#10;&#10;$\dfrac{{273.16}}{{200}} \times {T_A} = \dfrac{{273.16}}{{350}} \times {T_B}$&#10;&#10;Hence the ratio of the ${T_A}:{T_b} = 4:7$

Question 10

The temperature of a body on Kelvin scale is found to be $X\ K$. When it is measured by a Fahrenheit thermometer, it is found to be $X^o F$. Then $X$ is

Accepted Answer

We know, $F=1.8T-459.67$Where, $F$=Fahrenheit Temperature$T$=Kelvin TemperatureNow, $F=X&#176;F$ and $T=XF$, putting there we get,$X=1.8X-459.67$$\therefore X=\cfrac { 459.67 }{ 0.8 } \ =574.25\$

Revise with Concepts

Scales of Temperature

Relation Between Units of Temperature

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Quick Summary With Stories

Relation Between Different Scales of Temperature

Conversion of Temperature

Temperature Scales-II

At what temperature, the Celsius and Fahrenheit scales will read the same ?

A faulty thermometer has its fixed points marked $5^{o}$ and $95^{o}$ . This thermometer reads the temperature of a body as $59^{o}$ . Then the correct temperature on Celsius scale is

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

The freezing point on a thermometer is marked as $2 0^{o}$ and the boiling point as $15 0^{o}$ . A temperature of $6 0^{o} C$ on this thermometer will be read as

In terms of Charles' law, explain why - 273 $^{\circ} C$ is the lowest possible temperature.

If a thermometer reads freezing point of water as $2 0^{\circ} C$ and boiling point at $15 0^{\circ} C$ how much thermometer read when the actual temperature is $6 0^{\circ} C$ .

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39W and 239W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39C on the Celsius scale ?

At what temperature, the Fahrenheit scale reading is double of Celsius Scale reading?

Two absolute scales $A$ and $B$ have triple points of water defined to be $200 A$ and $300 B$ . A temperature is measured on these scales as $T_{A}$ and $T_{B}$ What is the relation between $T_{A}$ and $T_{B}$ ? Given, that triple point of water as $273.16 K$ .

The temperature of a body on Kelvin scale is found to be $X K$ . When it is measured by a Fahrenheit thermometer, it is found to be $X^{o} F$ . Then $X$ is

Revise with Concepts

Scales of Temperature

Relation Between Units of Temperature

Learn with Videos

Quick Summary With Stories

Relation Between Different Scales of Temperature

Conversion of Temperature

Temperature Scales-II

At what temperature, the Celsius and Fahrenheit scales will read the same ?

A faulty thermometer has its fixed points marked 5o and 95o. This thermometer reads the temperature of a body as 59o. Then the correct temperature on Celsius scale is

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

The freezing point on a thermometer is marked as 20o and the boiling point as 150o. A temperature of 60oC on this thermometer will be read as

In terms of Charles' law, explain why - 273 ∘C is the lowest possible temperature.

If a thermometer reads freezing point of water as 20∘C and boiling point at 150∘C how much thermometer read when the actual temperature is 60∘C.

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39W and 239W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39C on the Celsius scale ?

At what temperature, the Fahrenheit scale reading is double of Celsius Scale reading?

Two absolute scales A and B have triple points of water defined to be 200A and 300B . A temperature is measured on these scales as TA​ and TB​ What is the relation between TA​ and TB​ ? Given, that triple point of water as 273.16K .

The temperature of a body on Kelvin scale is found to be X K. When it is measured by a Fahrenheit thermometer, it is found to be XoF. Then X is

A faulty thermometer has its fixed points marked $5^{o}$ and $95^{o}$ . This thermometer reads the temperature of a body as $59^{o}$ . Then the correct temperature on Celsius scale is

The freezing point on a thermometer is marked as $2 0^{o}$ and the boiling point as $15 0^{o}$ . A temperature of $6 0^{o} C$ on this thermometer will be read as

In terms of Charles' law, explain why - 273 $^{\circ} C$ is the lowest possible temperature.

If a thermometer reads freezing point of water as $2 0^{\circ} C$ and boiling point at $15 0^{\circ} C$ how much thermometer read when the actual temperature is $6 0^{\circ} C$ .

Two absolute scales $A$ and $B$ have triple points of water defined to be $200 A$ and $300 B$ . A temperature is measured on these scales as $T_{A}$ and $T_{B}$ What is the relation between $T_{A}$ and $T_{B}$ ? Given, that triple point of water as $273.16 K$ .

The temperature of a body on Kelvin scale is found to be $X K$ . When it is measured by a Fahrenheit thermometer, it is found to be $X^{o} F$ . Then $X$ is