Question

Solve the following problems:
a. Equal heat is given to two objects A and B of mass 1 g. Temperature of A increases by 3°C and B by 5°C. Which object has more specific heat? And by what factor?
b. Liquid ammonia is used in ice factory for making ice from water. If water at 20°C is to be converted into 2 kg ice at 0°C, how many grams of ammonia are to be evaporated? (Given: The latent heat of vaporization of ammonia= 341 cal/g)
c. A thermally insulated pot has 150 g ice at temperature 0°C. How much steam of 100°C has to be mixed to it, so that water of temperature 50°C will be obtained? (Given : latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
d. A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg °C. It contains 250 gm of liquid at 30°C having specific heat of 0.4 kcal/kg °C. If we drop a piece of ice of mass 10 g at 0°C, What will be the temperature of the mixture?

Answer 1

a. Specific heat capacity of a body is given as
$s=\frac{∆Q}{m∆T}$
Let Q cal of heat is given to both A and B.
For body A,
$s_1=\frac{Q}{1\times 3}=\frac{Q}{3}\mathrm{cal}{\mathrm{g}}^{-1}{}^{\mathrm{o}}{\mathrm{C}}^{-1}$
For body B,
$s_2=\frac{Q}{1\times 5}=\frac{Q}{5}\mathrm{cal}{\mathrm{g}}^{-1}{}^{\mathrm{o}}{\mathrm{C}}^{-1}$
Now,
$$\begin{gathered} \frac{s_1}{s_2}=\frac{\frac{Q}{3}}{\frac{Q}{5}}=\frac{5}{3} \\ \Rightarrow s_1=\frac{5}{3}{s}_2 \end{gathered}$$
Thus, specific heat capacity of body A is more than body B and by a factor of $\frac{5}{3}$.

b. Amount of heat energy released in cooling 2 kg water from 20°C to 0°C = $2\times 1000\times 1\times 20=40000\mathrm{cal}$
Amount of heat energy released in converting 2 kg water at 0°C to ice = $2\times 1000\times 80=160000\mathrm{cal}$
Thus, total energy required in converting water at 20°C to ice = 200000 cal
Grams of ammonia to be evaporated = $\frac{200000}{341}=$586.5 g

c. Amount of heat required in converting 150 g ice to 0°C to water at 0°C = $150\times 80=12000\mathrm{cal}$
Amount of heat energy required in heating 150 g water at 0°C to 150 g water at 50°C = $150\times 1\times 50=7500\mathrm{cal}$
Total heat energy required to convert 150 g ice at 0°C to water at 50°C = 19500 cal
Let m g be the amount of steam be mixed with water to bring the final temperature of system at 50°C.
The amount of heat released in converting m g of steam at 100°C to water at 100°C = $m\times 540=540m$
The amount of heat released in converting m g of water at 100°C to water at 50°C = $m\times \times 1\times 50=50m$
Total heat energy released to convert m g steam at 100°C to water at 50°C = 590m cal
Using the principle of calorimetry, we have
590m = 19500
$m=\frac{19500}{590}=33\mathrm{g}$

d. Let the final temperature of the mixture be T.
Amount of heat required in converting 10 g ice to 0°C to water at 0°C = $10\times 80=800\mathrm{cal}$
Total amount of heat required in converting 10 g water to 0°C to water at T°C = $10\times 1\times T=10T$
Total heat energy required to convert 10 g ice at 0°C to water at T°C = 800 + 10T
Amount of heat released to raise the temperature of calorimeter at 30°C to T°C = $100\times 0.1\times (30-T)=10(30-T)$
Amount of heat released to raise the temperature of 250g of eater at 30°C to T°C = $250\times 0.4\times (30-T)=100(30-T)$
Total amount of heat released in the process = $110(30-T)$
Using the principle of calorimetry, we have
$110(30-T)$ = 800 + 10T
$\Rightarrow T=20.83{}^{\mathrm{o}}\mathrm{C}$