Question

10. Calculate the heat required to convert 0.6 kg of ice
at -20°C, kept in a calorimeter to steam at 100°C at
atmospheric pressure. Given the specific heat
capacity of ice = 2100 Jkg-K, specific heat
capacity of water is 4186 Jkg ? K-?, latent heat of
ice = 3.35 x 10 Jkg-1, and latent heat of steam =
2.256 x106 Jkg–1 [Delhi 08] (Ans. 1.8 <10% ]).​

Answer 1

Given :

Mass of ice = 0.6 kg

Temperature of ice = -20°C.

Temperature of steam = 100°C.

To find :

Heat required.

Solution :

When a substance is in a same phase but changing its temperature. Then because of change of temperature the required heat is equal to :

$H = mS (T_2 - T_1) \: \\ H = mS \Delta T \\ \: where\\ T_1 = \textbf{initial \: temperature } \\ T_2 \: = \textbf{final \: temperature }</p><p></p><p>$

(equation 1)

S = Specific heat of phase.

and

When a substance changes its phase,

then there is an amount of heat required to change its phase.

This heat required to phase change is equal to :

$H = mL \:$

(equation 2)

where

L = Latent heat of phase change.

We know the constant values in this case,

$\textbf{Specific heat capacity of ice = 2100 Jkg-K} \\ </strong></p><p><strong>[tex] \textbf{Specific heat capacity of ice = 2100 Jkg-K} \\ \textbf{Specific heat capacity of water = 4186 Jkg-K} \\ </strong></p><p><strong>[tex] \textbf{Specific heat capacity of ice = 2100 Jkg-K} \\ \textbf{Specific heat capacity of water = 4186 Jkg-K} \\ \textbf{Latent heat of ice = }3.35 \times 10 ^{5} Jkg ^{ - 1} \\ </strong></p><p><strong>[tex] \textbf{Specific heat capacity of ice = 2100 Jkg-K} \\ \textbf{Specific heat capacity of water = 4186 Jkg-K} \\ \textbf{Latent heat of ice = }3.35 \times 10 ^{5} Jkg ^{ - 1} \\ \:$

$\textbf{Latent heat of steam 2.256 M Jkg–1}$

To change ice into steam,

Firstly its temperature will increase from -20°C to 0°C.

so from equation 1,

Heat required for this process will be :

$H_1 = 0.6 \times 2100 \times (0 -(-20) \\ H_1 = 0.6 \times 2100 \times 20 \\ H_1 = 25200 \: \bf joules$

(step 1)

by 25200 joules heat, temperature of ice will reach to 0°C,

now to change the phase,from Ice to water,

The heat required will be (by equation 2) :

$H_2 = 0.6 \times 3.35 \times 10^5 \\ H_2 = 2.01 \times {10}^{5} \bf joules$

(step 2)

by 2.01 M joule heat, ice will change in water.

now,

we have to change the temperature of water from 0°C to 100°C.

So similar to step 1,

$H_3 = 0.6 \times 4186 \times (100 -0 \\ H_3 = 60\times 4186 \\ H_3 = 251160 \: \bf joules$

(step 3)

by 251560 joules of heat temperature of water raise to 100°C.

Now similar to step 2,

We have to change the phase from water to steam.

so,

$H_4 = 0.6 \times 2.256 \times {10}^{6} \\ H_4 =1.3536 \times {10}^{6} \bf joules$

(step 4)

so, by 1.3536 M joule heat, we can make all water into steam.

so,

Total heat required to change ice at -20°C to steam at 100°C.

$\bf \: H = H_1+ H_2 + H_3 + H_4$

So, Total heat required H is :

$\bf \: H = 25200 + 201000 + 251160 + 1353600 \: joules$

$\bf \: H = 1830960 \: joules$

So, Total heat required = 1,830,960 joules.