Calorimeter - Definition, Diagrams, Practice problems, FAQs

Nitin makes a cup of hot coffee in the morning. The cup of coffee is so hot that vapors are seen coming out from it. He then drops a piece of ice cube into a cup containing boiling coffee. He notices that the ice cube dissolves fast. He drops one more cube, and notices that the same thing happens again. Also, the vapors coming from the coffee have subsided. This is proof that coffee has cooled down. On the other hand, the ice cube would have gained heat. The same happens when two liquids having uneven temperatures are mixed; the hotter liquid becomes colder, and the colder liquid becomes hotter. Study of such heat lost or gained is calorimetry. In this article, we will explore calorimetry in detail.

Table of contents

• Principle of calorimetry
• Specific heat capacity
• Latent heat of fusion
• Examples of calorimetry
• Practice problems
• FAQs

Principle of calorimetry

Calorimetry deals with measurement of heat. When two bodies, having different temperatures, are placed in contact with each other, heat flows from the body having higher temperature to the one having lower temperature. Hence, the heat lost by the hotter body is equal to the heat lost by the cold body. This is in accordance with the law of conservation of energy. An apparatus used to measure the heat changes occuring in a reaction is called bomb calorimeter.

Specific heat capacity

The amount of heat required to raise the temperature of 1 kg of a substance by unity (1⁰C or 1K) is called specific heat capacity(C). If $\Delta \theta$ be the change in temperature, and Q be the amount of heat supplied to it. Then

$Q=mC\Delta \theta$; m- mass of the substance being heated.

The unit of specific heat capacity is $\frac{\frac{J}{kg}}{K}$ or $\frac{J}{kg{\frac{}{}}^{0}C}.$. Its dimensional formula is $\left[M^0{L}^2{T}^{-2}{\theta }^{-1}\right].$

Latent heat of fusion

When a substance undergoes a phase change, then some amount of heat is lost or gained. For instance, when ice melts to form water, it is an endothermic reaction, meaning heat is absorbed in it. On the other hand, when water freezes to form ice, heat is released. Let Q be the amount of heat released or absorbed, and L be the latent heat of fusion.Then,

Q=mL.

The CGS unit of latent heat of fusion is $\frac{cal}{g}$ and its SI unit is $Jk{g}^{-1}$. Its dimensional formula is $\left[M^0 L^2 T^{-2}\right].$

L_v- latent heat of fusion of steam$=540\frac{cal}{g}$

L_ice- latent heat of fusion of ice$=80\frac{cal}{g}$

Examples of calorimetry

Consider m1 kg of steam at 100⁰C being mixed with m₂ kg of water at 30⁰C. If T indicates the temperature of the mixture, then

Heat lost by steam=heat gained by water

$m_1{L}_v+m_1{C}_1\left(100-T\right)=m_2{C}_2\left(T-30\right)$

C₂- specific heat capacity of water

C₁- specific heat capacity of steam

Consider m₁ kg of ice at 0⁰C being mixed with m₂ kg of water at 40⁰C. If T indicates the common temperature of the mixture after mixing, then

$m_1{L}_{ice}+m_1{C}_3\left(T-0\right)=m_2 C_2\left(40-T\right)$

Heat gained by ice=heat lost by water

C₂- specific heat capacity of water.

C₃- specific heat capacity of ice.

Practice problems

Q1.Calculate the energy gained when 5 kg of water at 20⁰C is heated to its boiling point?(given, specific heat of water$=4.2\frac{kJ}{kg{\frac{}{}}^{0}C}.$

Answer. Given, m=5 kg, $\Delta \theta ={100}^{0}C-{20}^{0}C={80}^{0}C.$

$C=4200\frac{J}{kg{\frac{}{}}^{0}C}.$

$Q=mC\Delta \theta =5\times 4200\times 80=1680 kJ.$

Q2.100 kg of water at 80⁰ C is mixed with 300 kg of water at 60⁰ C. The final temperature reached by the mixture would be

(a)70⁰C                                                         (b)65⁰C
(c)60⁰C                                                        (d)75⁰C

Answer. b

Let T indicate the final temperature of the mixture.

$heat lost by water=100\times S_{water}\times \left(80-T\right)$

$heat gained by water=300\times S_{water}\times \left(T-60\right)$

$heat lost=heat gained$

$100\times S_{water}\times \left(80-T\right)=300\times S_{water}\times \left(T-60\right)$

$80-T=3T-180\Rightarrow 4 T={260}^0$

$T=6{5 }^0 C.$

Q3. 50g of ice at 0⁰ C is mixed with 50 g of water at 60⁰ C, final temperature of the mixture would be
Answer.

When 50 g of water at 60⁰C mixes with 50 g of ice at 0⁰ C.

Heat lost by water, $Q_1=m s \Delta \theta =50 \times 1\times 60=3000 cal.$

Q₂- heat absorbed by ice. Let m' be the mass of the ice.

$Q_2=m^{\text{'}{L}_{ice}}$;$L_{ice}=80\frac{cal}{g}$

$m^{\text{'}}=\frac{3000}{80}=37.5 g$

all of the ice has not melted. Hence the final temperature of the mixture would be 00C.

Q4. 100 grams of ice, initially at 0⁰C is mixed with 100 g of water kept at 100⁰ C. Find the final temperature of the mixture.

(a)10⁰C                                      (b)20⁰C
(c)30⁰C                                     (d)0⁰C

Answer. a

Heat lost by ice=Heat gained by water

$100\times 80+100\times 1\times \left(T-0\right)=100\times 1\times \left(100-T\right)$

$T={10}^{0}C.$

Q4.Calculate the heat capacity of 40 g of aluminum. (Specific heat capacity of aluminum$C=0.2\frac{cal}{g{\frac{}{}}^{0}C}$

(a)$40 cal{\frac{}{}}^{0}C$                                                   (b)$160 cal{\frac{}{}}^{0}C$
(c)$200 cal{\frac{}{}}^{0}C$                                                 (d)$8 cal{\frac{}{}}^{0}C$

Answer. d

Mass of the aluminum m=40 g, Specific heat of aluminum $C=0.2\frac{cal}{g{\frac{}{}}^{0}C}$

Heat capacity$=m \times c=40\times 0.2=8 cal{\frac{}{}}^{0}C.$

FAQs

Q1.What are some limitations of calorimetry?
Answer. If proper insulation is not provided, the bomb calorimeter loses heat to the surroundings. Also, uneven heating occurs if the mixture is not well mixed. Explosion of substances like TNT cannot be contained in a calorimeter.

Q2.On what principle does a calorimeter operate?
Answer. Calorimeter works on the principle of conservation of energy. When a mixture of ice and water is taken into the calorimeter, the heat lost by water is equal to the heat gained by ice. When a mixture of steam and ice is taken, heat lost by steam is equal to the heat gained by the ice.

Q3.Why is water used as radiators?
Answer. Water is used in car radiators because of its high specific heat capacity. This is the reason why water heats up slower compared to sand in desert areas. Due to its high specific heat capacity, it takes a lot more time to heat up than sand.

Q4.What are the factors affecting specific heat capacity?
Answer. Specific heat capacity is the heat capacity of the material per unit mass. The heat transferred to a system depends upon-the mass of the system, the change in phase it undergoes, and the temperature change in the system.