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11th
Physics
Thermal Properties of Matter
Calorimetry
$Calorimetry Questions and A...$

Important questions on Calorimetry

Q: How much heat transfer is required to raise the temperature of a 0.45kg aluminium pot containing 1.5kg of ice from -10^{o}C to boiling point and then boil away the water? How long does it take if the rate of heat transfer is 300\:W ?

Initial temperature of pot + ice= -10°C Heat required to convert ice from,-10°C to 0°C= mc \Delta T= Q_1 Where,m=MassC=Thermal capacity of ice \Delta T=Temperature differenceThermal capacity of ice= 0.5cal/g°C Q_1=mc\Delta T=1500\times\dfrac{1}{2}\times10=7,500 caloriesHeat required to change the state of ice =Q_2 =mLWhere, m=Mass, L=Latent heat Q_2=1500\times80=1,20,000 caloriesHeat required to change the temperature of water(melted ice) =Q_3=mc\Delta T Q_3=1500\times1\times100=1,50,000 caloriesHeat required to change the state of water= mL= Q_4 Q_4=1500\times540 = 8,10,000 caloriesHeat required to change the temperature of aluminum pot =Q_5=mc\Delta T Specific heat capacity of aluminum =0.21cal/g°C Q_5=mc\Delta T 450\times0.21\times110 = 10,395 caloriesTotal heat required =Q_1+Q_2+Q_3+Q_4+Q_5 = 10,97,895 calories = 46,11,159 JoulesWe are given P =300W=300 Joules\: per\: second 300 Joules are produced in = 1 s1 Joule is produced in =\dfrac{1}{300} s46,11,159 Joules is produced in =\dfrac{1}{300}\times46,11,159 = 15,370.53 seconds

Q: The diameter of a solid metallic sphere is equal to the inner diameter of a hollow metallic sphere. If inner diameter of hollow sphere is two thirds of its outer diameter. Both the spheres are made of same metal. If an equal amount of heat is supplied the both the spheres. Find the ratio of increase in temperature of the solid sphere to that of the hollow one?

Given: The diameter of a solid metallic sphere is equal to the inner diameter of a hollow metallic sphere. If inner diameter of hollow sphere is two thirds of its outer diameter. Both the spheres are made of same metal. If an equal amount of heat is supplied the both the spheres. To find the ratio of increase in temperature of the solid sphere to that of the hollow oneSolution:We know that,Heat transfer = Q= mc\Delta T Where, m=mass, c=specific heat capacity, \Delta T =temperature. Q=\rho Vc\Delta T (where \rho =density, V=volume, as m=\rho V )Now if there are two bodies to which equal amount of heat supplied and are of same metal. Q_1=\rho_1V_1c_1\Delta T_1 Q_2=\rho_2V_2c_2\Delta T_2 If same metal (\rho_1=\rho_2, c_1=c_2) Ratio of increase in temperature \dfrac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } =\dfrac { { V }_{ 2 } }{ { V }_{ 1 } } .....(i) Now in question,Let the diameter of solid sphere be = d Inner diameter of hollow sphere = d Outer diameter of hollow sphere = \dfrac{3}{2}d temperature rise in solid sphere = \Delta T_1 temperature rise in hollow sphere= \Delta T_2 According to equation (i) \dfrac { \Delta { T }_{ 1 } }{ \Delta { T }_{ 2 } } =\dfrac { { V }_{ 2 } }{ { V }_{ 1 } } Volume of hollow sphere =\dfrac { 4 }{ 3 } \pi \left( { r }_{ 2 }^{ 3 }-{ r }_{ 1 }^{ 3 } \right) r_2 =outer radius r_1 = inner radiusVolume of solid sphere =\dfrac{4}{3}\pi r^3 \dfrac { { \Delta T }_{ 1 } }{ { \Delta T }_{ 2 } } =\dfrac { \dfrac { 4\pi }{ 3 } \left[ \left( \dfrac { 27d^3}{ 64 } -\dfrac { { d }^{ 3 } }{ 8 } \right) \right] }{ \dfrac { 4\pi }{ 3 } \left[ \dfrac { { d }^{ 3 } }{ 8 } \right] } =\dfrac { \dfrac { 27-8 }{ 64 } }{ \dfrac { 1 }{ 8 } } \\\implies \dfrac { { \Delta T }_{ 1 } }{ { \Delta T }_{ 2 } } =\dfrac {19}{64}\times 8\\\implies \dfrac { { \Delta T }_{ 1 } }{ { \Delta T }_{ 2 } } =\dfrac {19}8 is the the ratio of increase in temperature of the solid sphere to that of the hollow one

Q: 5gm of steam at 100^oC is passed into calorimeter containing liquid , Temperature of liquid rises from 32^oC to 40^oC . Then water equivalent of calorimeter and content is

Latent heat of vaporization of water = 2260kJ/Kg The specific heat capacity of water = 4185.5 J/Kg Heat lost by steam = heat gained by the water and calorimeter Formula : Heat gained by water = mcФ m = mass of water c = specific heat capacity of water. Ф = Change in temperature. = 40 - 32 = 8 Heat lost by steam = mLv + mcФ Lv = latent heat of vaporisation m = mass of steam. Ф = 100 - 40 = 60 Doing the substitution : 2260000 \times 0.005 + 60 \times 0.005 \times 4185.5 = m \times 4185.5 \times 8 12555.65 = 33484m m = \dfrac {12555.65} {33484} = 0.37497 kg = 0.37497 \times 1000 = 374.97 kg = 374.97kg

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A copper calorimeter can with a mass of $0.446 k g$ contains $0.0950 k g$ of ice . The system is initially at $0 . 0^{\circ} C .$ (a) If $0.0350 k g$ of steam at $100 . 0^{\circ} C$ and $1.00 a t m$ pressure is added to the can , what is the final temperature of the calorimeter can and its contents ? (b) At the final temperature , how many kilograms are there of ice , how many of liquid water , and how many of steam ?

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A $200 - g$ aluminium calorimeter contains contains $500 g$ of water at $2 0^{\circ} C$ . A $100 - g$ piece of ice cooled to $- 2 0^{\circ} C$ is placed in the calorimeter .(a) Find the final temperature of the system ,assuming no heat loss . (Assume that the specific heat of ice is $2.0 k J / k g K$ .) (b) A second $200 g$ piece of ice at $- 2 0^{\circ} C$ is added . How much ice remains in the system after it reaches equilibrium ? (c) Would you give a different answer for (b) if both pieces of ice were added at the same time ?

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How much heat transfer is required to raise the temperature of a $0.45 k g$ aluminium pot containing $1.5 k g$ of ice from $- 1 0^{o} C$ to boiling point and then boil away the water? How long does it take if the rate of heat transfer is $300 W$ ?

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A $200 g$ aluminium calorimeter contains $500 g$ of water at $2 0^{\circ} C .$ An aluminium shot of mass $300 g$ is heated to $10 0^{\circ} C$ and is then placed in the calorimeter . (a) Using the value of the specific heat of aluminium given in table find the final temperature of the system , assuming that no heat is lost to the surroundings . (b) The error due to heat transfer between the system and its surroundings can be minimized if the initial temperature of the water and calorimeter is chosen to be $Δ t_{w}$ below room temperature , where $Δ t_{w}$ is the temperature change of the calorimeter and water during the measurement . Then the final temperature is $Δ t_{w}$ above room temperature . What should be the initial temperature of the water and container if the room temperature is $2 0^{\circ} C ?$

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Utensils used for efficient cooking should have

Large heat capacity

Small heat capacity

Medium heat capacity

Any heat capacity

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The diameter of a solid metallic sphere is equal to the inner diameter of a hollow metallic sphere. If inner diameter of hollow sphere is two thirds of its outer diameter. Both the spheres are made of same metal. If an equal amount of heat is supplied the both the spheres. Find the ratio of increase in temperature of the solid sphere to that of the hollow one?

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The water equivalent of a 400 g copper calorimeter (specific heat =0.1 cal/ $g^{o} C$ )-

$40 g$

$4000 g$

$200 g$

$4 g$

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$5 g m$ of steam at $10 0^{o} C$ is passed into calorimeter containing liquid , Temperature of liquid rises from $3 2^{o} C$ to $4 0^{o} C$ . Then water equivalent of calorimeter and content is

$40$ gram

$375$ gram

$300$ gram

$160$ gram

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A flow calorimeter is an apparatus used to measure the specific heat of a liquid . The technique of flow calorimeter involves measuring the temperature difference between the input and output points of a flowing stream of the liquid while energy is added by heat at a known rate . In one particular experiment , a liquid with a density of $0.780 g / c m^{3}$ flows through the calorimeter at the rate of $4.00 c m^{3} / s$ At steady state , a temperature difference of $4.8 0^{\circ} C$ is established between the input and output points when energy is supplied by heat at the rate of $30.0 J /$ . What is the specific heat of the liquid ?

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A vessel contains M grams of water at a certain temperature and water at a certain other temperatures is passed into it at a constant rate of m g/s. The variation of temperature of the mixture with time is shown in Fig. 1.90. The values of M and m are, respectively (the heat exchanged after a long time is 800 cal)

Important questions on Calorimetry

How much heat transfer is required to raise the temperature of a $0.45 k g$ aluminium pot containing $1.5 k g$ of ice from $- 1 0^{o} C$ to boiling point and then boil away the water? How long does it take if the rate of heat transfer is $300 W$ ?

Utensils used for efficient cooking should have

Large heat capacity

Small heat capacity

Medium heat capacity

Any heat capacity

The water equivalent of a 400 g copper calorimeter (specific heat =0.1 cal/ $g^{o} C$ )-

$40 g$

$4000 g$

$200 g$

$4 g$

$5 g m$ of steam at $10 0^{o} C$ is passed into calorimeter containing liquid , Temperature of liquid rises from $3 2^{o} C$ to $4 0^{o} C$ . Then water equivalent of calorimeter and content is

$40$ gram

$375$ gram

$300$ gram

$160$ gram

40 & 2

40 & 4

20 & 4

20 & 2

More important questions

Important questions on Calorimetry

How much heat transfer is required to raise the temperature of a 0.45kg aluminium pot containing 1.5kg of ice from −10oC to boiling point and then boil away the water? How long does it take if the rate of heat transfer is 300W?

Utensils used for efficient cooking should have

Large heat capacity

Small heat capacity

Medium heat capacity

Any heat capacity

The water equivalent of a 400 g copper calorimeter (specific heat =0.1 cal/goC)-

40g

4000g

200g

4g

5gm of steam at 100oC is passed into calorimeter containing liquid , Temperature of liquid rises from 32oC to 40oC. Then water equivalent of calorimeter and content is

40 gram

375 gram

300 gram

160 gram

40 & 2

40 & 4

20 & 4

20 & 2

More important questions

How much heat transfer is required to raise the temperature of a $0.45 k g$ aluminium pot containing $1.5 k g$ of ice from $- 1 0^{o} C$ to boiling point and then boil away the water? How long does it take if the rate of heat transfer is $300 W$ ?

The water equivalent of a 400 g copper calorimeter (specific heat =0.1 cal/ $g^{o} C$ )-

$40 g$

$4000 g$

$200 g$

$4 g$

$5 g m$ of steam at $10 0^{o} C$ is passed into calorimeter containing liquid , Temperature of liquid rises from $3 2^{o} C$ to $4 0^{o} C$ . Then water equivalent of calorimeter and content is

$40$ gram

$375$ gram

$300$ gram

$160$ gram