Q: Three rods made of same material and having same cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at \$0^{\circ}C\$ and \$90^{\circ}C\$ respectively. The temperature of the junction of the three rods will be

Let \$\theta ^{\circ}C\$ be the temperature at \$B\$. Let \$Q\$ be the heat flowing per second from \$A\$ to \$B\$ on account of temperature difference by conductivity.\$\therefore Q = \dfrac {KA(90 - \theta)}{l} ... (i)\$where \$k =\$ thermal conductivity of the rod, \$A =\$ Area of cross section of the rod, \$l =\$ length of the rod. By symmetry, the same will be the case for heat flow from \$C\$ to \$B\$.\$\therefore\$ The heat flowing per second from \$B\$ to \$D\$ will be\$2Q = \dfrac {KA(\theta - 0)}{l} .... (ii)\$Dividing Eq. (ii) by Eq. (i)\$2 = \dfrac {\theta}{90 - \theta} \Rightarrow \theta = 60^{\circ}C\$.

Q: A body cools from \${80}^{\circ}\$C to \${50}^{\circ}\$C in \$5\$ minutes. Calculate the time it takes to cool from \${60}^{\circ}\$C to \${30}^{\circ}\$C. The temperature of the surroundings is \${20}^{\circ}\$C.

Here, Initial temperature \$( {T}_{i}) = {80}^{\circ}\$C Final temperature \$( {T}_{f}) = {50}^{\circ}\$C Temperature of the surrounding \$( {T}_{0}) = {20}^{\circ}\$C \$ t = 5\$ min According to Newton's law of cooling,Rate of cooling \$\dfrac{dT}{dt} = K[\dfrac{\left({T}_{i}+{T}_{f}\right)}{2}-{T}_{o}]\$\$\dfrac{\left({T}_{f}-{T}_{i}\right)}{t}=K\left[\dfrac{\left(80 +50\right)}{2} - 20\right]\$\$\dfrac{80-50}{5} = K[ 65 - 20]\$\$6 = K\times 45\$\$K = \dfrac{6}{45} = \dfrac{2}{15}\$ In second condition, initial temperature \$={T}_{i}= {60}^{\circ}\$C Final temperature \${T}_{f}={30}^{\circ}\$C Time taken for cooling is \$t\$ According to Newton's law of cooling \$\dfrac{( 60 - 30)}{t} = \dfrac{2}{15} \left[ \dfrac{\left(60+30\right)}{2} -20\right]\$\$\dfrac{30}{t}=\dfrac{2}{15}\times 25\$\$\dfrac{30}{t} = \dfrac{50}{15}=\dfrac{10}{3}\$\$\therefore t = 9\$ min

Q: The temperature of equal masses of three different liquids \$A, B\$ and \$C\$ are \$12^{\circ}C, 19^{\circ}C\$ and \$28^{\circ}C\$ respectively. The temperature when \$A\$ and \$B\$ are mixed is \$16^{\circ}C\$ and when \$B\$ and \$C\$ are mixed it is \$23^{\circ}C\$. What should be the temperature when \$A\$ and \$C\$ are mixed?

Let \$m\$ be the mass of each liquid and \$S_{A}, S_{B}, S_{C}\$ be specific heats of liquids \$A, B\$ and \$C\$ respectively. When \$A\$ and \$B\$ are mixed. The final temperature is \$16^{\circ}C\$.\$\therefore\$ Heat gained by \$A =\$ heat lost by \$B\$i.e., \$mS_{A} = (16 - 12) = mS_{B} (19 - 16)\$i.e., \$S_{B} = \dfrac {4}{3}S_{A} ..... (i)\$When \$B\$ and \$C\$ are mixed. Heat gained by \$B =\$ Heat lost by \$C\$i.e., \$mS_{B} = (23 - 19) = mS_{C} (28 - 23)\$i.e., \$S_{C} = \dfrac {4}{5}S_{B} ..... (ii)\$From eq. (i) and (ii)\$S_{C} = \dfrac {4}{5}\times \dfrac {4}{3} S_{A} = \dfrac {16}{15}S_{A}\$When \$A\$ and \$C\$ are mixed, let the final temperature be \$\theta\$\$\therefore mS_{A} (\theta - 12) = mS_{C} (28 - \theta)\$i.e., \$\theta - 12 = \dfrac {16}{15} (28 - \theta)\$By solving, we get,\$\theta = \dfrac {628}{31} = 20.26^{\circ}C\$.

Q: A black body at a temperature of \$227^{\circ}C\$ radiates heat energy at the rate \$5\;cal/cm^2/s\$. At a temperature of \$727^{\circ}C\$, the rate of heat radiated per unit area in \$cal/cm^2\$ will be:

Heat energy per unit time \$R = \sigma AeT^4\$Thus heat energy per unit area \$R' = \sigma e T^4\$\$\implies\$ \$\dfrac{R'_2}{R'_1} = \bigg( \dfrac{T_2}{T_1} \bigg)^4\$Initial temperature of the body \$T_1 =227 ^oC = (227+ 273) K = 500 K\$Final temperature of the body \$T_2 =727 ^oC = (727+ 273) K = 1000 K\$\$\therefore\$ \$\dfrac{R'_2}{5} = \bigg( \dfrac{1000}{500} \bigg)^4\$\$\dfrac{R'_2}{5} = 16\$ \$\implies R'_2 = 80\$ \$cal/cm^2\$

Q: A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at \$27.0 ^oC\$. What is the change in the diameter of the hole when the sheet is heated to \$227^oC \$? Coefficient of the linear expansion of copper \$=1.70\times 10^{-5}K^{-1} \$

Initial temperature, \$T_1=27.0^oC\$Diameter of the hole at \$T_1, d_1=4.24cm\$Final temperature, \$T_2=227^oC\$Diameter of the hole at \$T_2=D_2\$Co-efficient of linear expansion of copper, \$\alpha_{Cu} =1.70\times 10^{-5}K^{-1}\$For co-efficient of superficial expansion β,and change in temperature \$\triangle T\$, we have the relation:Change in area \$(\triangle)\$ / Original area (A) = \$\beta \triangle T\$\$[(\pi d_2^2/4)-(\pi d_2^2/4)]/(\pi d_1^1/4)=\triangle A/A\$\$\therefore \triangle A/A = (d_2^2-d_1^2)/d_1^2\$But \$\beta = 2\alpha\$\$\therefore (d_2^2-d_1^2)/d_1^2=2\alpha\triangle T\$\$(d_2^2/d_1^2)-1=2\alpha(T_2-T_1)\$\$d_2^2/4.24^2=2\times 1.7\times 10^{-5}(227-27)+1\$\$d_2^2=17.98\times 1.0068=18.1\$\$\therefore d_2=4.2544 cm\$Change in diameter \$=d_2-d_1=4.2544-4.24=0.0144 cm\$Hence, the diameter increases by \$1.44\times 10^{-2}cm\$

Q: Two rods of length \$l_1\$ and \$l_2\$ are made of materials whose coefficients of linear expansion are \$\alpha_1\$ and \$\alpha_2\$. If the difference between their lengths is independent of temperature, then :

Length of Rods \$=L_{1}\$ & \$L_{2}\$Coefficient of thermal Expansion \$=\alpha_{1}\$ & \$\alpha_{2}\$According to the question\$\triangle L_{1}=\triangle L_{2}\$\$L_{1}\alpha_{1}\triangle T=L_{2}\alpha_{2}\triangle T\$\$\dfrac{L_{1}}{L_{2}}=\dfrac{\alpha_{2}}{\alpha_{1}}\$

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: Six identical conducting rods are joined as shown in figure. Points \$A\$ and \$D\$ are maintained at temperature of \$200 ^\circ C\$ and \$20 ^\circ C\$ respectively. The temperature of junction \$B\$ will be :

\${ k }_{ BC }=k\$ (wheat stone bridge)\$ \dfrac { { k }A(200-{ t }_{ B }) }{ l } =\dfrac { kA({ t }_{ B }-{ t }_{ C }) }{ l } =\dfrac { kA({ t }_{ C }-20) }{ l } \$\$ { t }_{ B }={ 140 }^{ \circ }C{ \quad \& \quad t }_{ C }={ 80 }^{ \circ }C\$

Question 1

Three rods made of same material and having same cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at $0^{\circ}C$ and $90^{\circ}C$ respectively. The temperature of the junction of the three rods will be

Accepted Answer

Let $\theta ^{\circ}C$ be the temperature at $B$. Let $Q$ be the heat flowing per second from $A$ to $B$ on account of temperature difference by conductivity.$\therefore Q = \dfrac {KA(90 - \theta)}{l} ... (i)$where $k =$ thermal conductivity of the rod, $A =$ Area of cross section of the rod, $l =$ length of the rod. By symmetry, the same will be the case for heat flow from $C$ to $B$.$\therefore$ The heat flowing per second from $B$ to $D$ will be$2Q = \dfrac {KA(\theta - 0)}{l} .... (ii)$Dividing Eq. (ii) by Eq. (i)$2 = \dfrac {\theta}{90 - \theta} \Rightarrow \theta = 60^{\circ}C$.

Question 2

A body cools from ${80}^{\circ}$C to ${50}^{\circ}$C in $5$ minutes. Calculate the time it takes to cool from ${60}^{\circ}$C to ${30}^{\circ}$C. The temperature of the surroundings is ${20}^{\circ}$C.

Accepted Answer

Here, Initial temperature $( {T}_{i}) = {80}^{\circ}$C Final temperature $( {T}_{f}) = {50}^{\circ}$C Temperature of the surrounding $( {T}_{0}) = {20}^{\circ}$C $ t = 5$ min According to Newton's law of cooling,Rate of cooling $\dfrac{dT}{dt} = K[\dfrac{\left({T}_{i}+{T}_{f}\right)}{2}-{T}_{o}]$$\dfrac{\left({T}_{f}-{T}_{i}\right)}{t}=K\left[\dfrac{\left(80 +50\right)}{2} - 20\right]$$\dfrac{80-50}{5} = K[ 65 - 20]$$6 = K\times  45$$K = \dfrac{6}{45} = \dfrac{2}{15}$ In second condition, initial temperature $={T}_{i}= {60}^{\circ}$C Final temperature ${T}_{f}={30}^{\circ}$C Time taken for cooling is $t$ According to Newton's law of cooling $\dfrac{( 60 - 30)}{t} = \dfrac{2}{15} \left[ \dfrac{\left(60+30\right)}{2} -20\right]$$\dfrac{30}{t}=\dfrac{2}{15}\times  25$$\dfrac{30}{t} = \dfrac{50}{15}=\dfrac{10}{3}$$\therefore t = 9$ min

Question 3

The temperature of equal masses of three different liquids $A, B$ and $C$ are $12^{\circ}C, 19^{\circ}C$ and $28^{\circ}C$ respectively. The temperature when $A$ and $B$ are mixed is $16^{\circ}C$ and when $B$ and $C$ are mixed it is $23^{\circ}C$. What should be the temperature when $A$ and $C$ are mixed?

Accepted Answer

Let $m$ be the mass of each liquid and $S_{A}, S_{B}, S_{C}$ be specific heats of liquids $A, B$ and $C$ respectively. When $A$ and $B$ are mixed. The final temperature is $16^{\circ}C$.$\therefore$ Heat gained by $A =$ heat lost by $B$i.e., $mS_{A} = (16 - 12) = mS_{B} (19 - 16)$i.e., $S_{B} = \dfrac {4}{3}S_{A} ..... (i)$When $B$ and $C$ are mixed. Heat gained by $B =$ Heat lost by $C$i.e., $mS_{B} = (23 - 19) = mS_{C} (28 - 23)$i.e., $S_{C} = \dfrac {4}{5}S_{B} ..... (ii)$From eq. (i) and (ii)$S_{C} = \dfrac {4}{5}\times \dfrac {4}{3} S_{A} = \dfrac {16}{15}S_{A}$When $A$ and $C$ are mixed, let the final temperature be $\theta$$\therefore mS_{A} (\theta - 12) = mS_{C} (28 - \theta)$i.e., $\theta - 12 = \dfrac {16}{15} (28 - \theta)$By solving, we get,$\theta = \dfrac {628}{31} = 20.26^{\circ}C$.

Question 4

A black body at a temperature of $227^{\circ}C$ radiates heat energy at the rate $5\;cal/cm^2/s$. At a temperature of $727^{\circ}C$, the rate of heat radiated per unit area in $cal/cm^2$ will be:

Accepted Answer

Heat energy per unit time          $R = \sigma AeT^4$Thus heat energy per unit area       $R'  = \sigma e T^4$$\implies$    $\dfrac{R'_2}{R'_1}  = \bigg( \dfrac{T_2}{T_1} \bigg)^4$Initial temperature of the body       $T_1   =227 ^oC   = (227+ 273) K  = 500  K$Final temperature of the body       $T_2   =727 ^oC   = (727+ 273) K  = 1000  K$$\therefore$     $\dfrac{R'_2}{5}  = \bigg( \dfrac{1000}{500} \bigg)^4$$\dfrac{R'_2}{5}  = 16$               $\implies   R'_2  = 80$   $cal/cm^2$

Question 5

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 6

A hole is drilled in a copper sheet.  The diameter of the hole is 4.24 cm at $27.0 ^oC$. What is the change in the diameter of the hole when the sheet is heated to $227^oC $? Coefficient  of the linear expansion of copper $=1.70\times 10^{-5}K^{-1} $

Accepted Answer

Initial temperature, $T_1=27.0^oC$Diameter of the hole at $T_1, d_1=4.24cm$Final temperature, $T_2=227^oC$Diameter of the hole at $T_2=D_2$Co-efficient of linear expansion of copper, $\alpha_{Cu} =1.70\times 10^{-5}K^{-1}$For co-efficient of superficial expansion &#946;,and change in temperature $\triangle T$, we have the relation:Change in area $(\triangle)$ /  Original area (A)  = $\beta \triangle T$$[(\pi d_2^2/4)-(\pi d_2^2/4)]/(\pi d_1^1/4)=\triangle A/A$$\therefore \triangle A/A = (d_2^2-d_1^2)/d_1^2$But $\beta = 2\alpha$$\therefore (d_2^2-d_1^2)/d_1^2=2\alpha\triangle T$$(d_2^2/d_1^2)-1=2\alpha(T_2-T_1)$$d_2^2/4.24^2=2\times 1.7\times 10^{-5}(227-27)+1$$d_2^2=17.98\times 1.0068=18.1$$\therefore d_2=4.2544 cm$Change in diameter $=d_2-d_1=4.2544-4.24=0.0144 cm$Hence, the diameter increases by $1.44\times 10^{-2}cm$

Question 7

Two rods of length $l_1$ and $l_2$ are made of materials whose coefficients of linear expansion are $\alpha_1$ and $\alpha_2$. If the difference between their lengths is independent of temperature, then :

Accepted Answer

Length of Rods $=L_{1}$ & $L_{2}$Coefficient of thermal Expansion $=\alpha_{1}$ & $\alpha_{2}$According to the question$\triangle L_{1}=\triangle L_{2}$$L_{1}\alpha_{1}\triangle T=L_{2}\alpha_{2}\triangle T$$\dfrac{L_{1}}{L_{2}}=\dfrac{\alpha_{2}}{\alpha_{1}}$

Question 8

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 9

Six identical conducting rods are joined as shown in figure. Points $A$ and $D$ are maintained at temperature of $200 ^\circ C$ and $20 ^\circ C$ respectively. The temperature of junction $B$ will be :

Accepted Answer

${ k }_{ BC }=k$ (wheat stone bridge)$ \dfrac { { k }A(200-{ t }_{ B }) }{ l } =\dfrac { kA({ t }_{ B }-{ t }_{ C }) }{ l } =\dfrac { kA({ t }_{ C }-20) }{ l } $$ { t }_{ B }={ 140 }^{ \circ }C{ \quad \& \quad t }_{ C }={ 80 }^{ \circ }C$

Question 10

A circular metal loop is heated at point $O$ as shown in above Figure.i) In which direction would heat flow in the loop?ii) In which order the pins at points P, Q and R fixed with the help of wax fall if points O, P, Q and R are equidistant from each other?

Accepted Answer

According to the given condition.i) The heat will flow in the direction from O to P and O to R.ii) Firstly the pins at R and P will fail concurrently followed by the pin at Q.