Call Now
1800-102-2727Nitin 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
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.
The amount of heat required to raise the temperature of 1 kg of a substance by unity (1^{0}C 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 $\left[{M}^{0}{L}^{2}{T}^{-2}{\theta}^{-1}\right].$
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 100^{0}C being mixed with m_{2} kg of water at 30^{0}C. If T indicates the temperature of the mixture, then
Heat lost by steam=heat gained by water
${m}_{1}{L}_{v}+{m}_{1}{C}_{1}\left(100-T\right)={m}_{2}{C}_{2}\left(T-30\right)$
C_{2}- specific heat capacity of water
C_{1}- specific heat capacity of steam
Consider m_{1} kg of ice at 0^{0}C being mixed with m_{2} kg of water at 40^{0}C. 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
C_{2}- specific heat capacity of water.
C_{3}- specific heat capacity of ice.
Q1.Calculate the energy gained when 5 kg of water at 20^{0}C is heated to its boiling point?(given, specific heat of water$=4.2\frac{kJ}{kg{\frac{}{}}^{0}C}.$
Answer. Given, m=5 kg, $\Delta \theta ={100}^{0}C-{20}^{0}C={80}^{0}C.$
$C=4200\frac{J}{kg{\frac{}{}}^{0}C}.$
$Q=mC\Delta \theta =5\times 4200\times 80=1680kJ.$
Q2.100 kg of water at 80^{0} C is mixed with 300 kg of water at 60^{0} C. The final temperature reached by the mixture would be
(a)70^{0}C (b)65^{0}C
(c)60^{0}C (d)75^{0}C
Answer. b
Let T indicate the final temperature of the mixture.
$heatlostbywater=100\times {S}_{water}\times \left(80-T\right)$
$heatgainedbywater=300\times {S}_{water}\times \left(T-60\right)$
$heatlost=heatgained$
$100\times {S}_{water}\times \left(80-T\right)=300\times {S}_{water}\times \left(T-60\right)$
$80-T=3T-180\Rightarrow 4T={260}^{0}$
$T=6{5}^{0}C.$
Q3. 50g of ice at 0^{0} C is mixed with 50 g of water at 60^{0} C, final temperature of the mixture would be
Answer.
When 50 g of water at 60^{0}C mixes with 50 g of ice at 0^{0} C.
Heat lost by water, ${Q}_{1}=ms\Delta \theta =50\times 1\times 60=3000cal.$
Q_{2}- heat absorbed by ice. Let m' be the mass of the ice.
${Q}_{2}={m}^{\text{'}{L}_{ice}}$;${L}_{ice}=80\frac{cal}{g}$
${m}^{\text{'}}=\frac{3000}{80}=37.5g$
all of the ice has not melted. Hence the final temperature of the mixture would be 00C.
Q4. 100 grams of ice, initially at 0^{0}C is mixed with 100 g of water kept at 100^{0} C. Find the final temperature of the mixture.
(a)10^{0}C (b)20^{0}C
(c)30^{0}C (d)0^{0}C
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)$40cal{\frac{}{}}^{0}C$ (b)$160cal{\frac{}{}}^{0}C$
(c)$200cal{\frac{}{}}^{0}C$ (d)$8cal{\frac{}{}}^{0}C$
Answer. d
Mass of the aluminum m=40 g, Specific heat of aluminum $C=0.2\frac{cal}{g{\frac{}{}}^{0}C}$
Heat capacity$=m\times c=40\times 0.2=8cal{\frac{}{}}^{0}C.$
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.