Question

A 10 kg block of ice is converted to water vapor. The block of ice starts at −20ºC;first, it must be completely melted, and then it is heated to 100ºC.Find the energy required to do this. Use a value of 2,108J/kgºC for the specific heat capacity of ice and use a value of 4,184J/kgºC for the specific heat capacity of water. Use a value of 334 kJ/kg for the latent heat of fusion of water and use a value of 2,258 kJ/kg for the latent heat of vaporization of water. Give your answer to 3 significant figures.

Answer 1

Given :

Mass of ice 10 kg.

Temperature kof ice = -20°C.

Temperature of steam = 100°C.

Specific heat capacity of ice = 2,108J/kgºC

Specific heat capacity of water = 4,184J/kgºC

Latent heat of fusion of water = 334 kJ/kg

Latent heat of vaporization of water = 2,258 kJ/kg

To find :

Energy required to covert all ice in vapour.

Solution :

When we change temperature of ice or water,

The energy need for changing temperature from temperature T1 to temperature T2 is

$H = ms\Delta T = ms(T_2 - T_1)$

(equation 1)

where

s = specific heat of ice or water.

m = mass of material.

∆T = Change in temperature.

H = Heat required.

When we change the phase of temperature,

like from ice to water or water to vapour,

The heat required :

$H = mL \:$

(equation 2)

So,

to change 10 kg of ice to vapour,

1. We will first change temperature of ice from -20°C to 0°C,

and find heat by equation 1,

2. Then change the phase of ice to water by equation2

3. Then increase the temperature of water from 0°C to 100°C by equation 1.

4. Then change the phase of water to vapour by equation 2.

So,

Total Energy required :

$H = (10 \times 2108 \times (0 - ( - 20))) + (10 \times 334) + (10 \times 4184 \times (100 - 0)) + (10 \times 2258)$

so,

$H = (10 \times 2108 \times 20) + (10 \times 334) + (10 \times 4184 \times 100) + (10 \times 2258)$

$H = (2108 \times 200) + 3340 + (4184 \times 1000)+ 22580$

So,

Total Energy required = 4,631,520 Joules.= 4.632 MJ

in 3 significant figures,

Total Energy required = 4.63 MJ.