Question

A container holding 1.2 kg of water at 20.0 °C is placed in a freezer that is kept at –20.0 °C. The water freezes and comes into thermal equilibrium with the interior of the freezer.
a) How much heat is extracted from the water in thisprocess?
b) What is the minimum amount of electrical energy required bythe refrigerator to carry out this process if it operates betweenreservoirs at temperatures of 20.0 °C and -20.0 °C?

Answer 1

Answer:

(a) $Q=556464\ J$

(b) 556464 joule

Explanation:

Given:

• mass of water, $m_w=1.2\ kg$

• initial temperature of water, $T_i=20^{\circ}C$

• final temperature of frozen water, $T_f=-20{}^{\circ}C$

The conversion of water of 20.0 °C to the ice of –20.0 °C will comprise of three steps:

1. cooling of water to 0 °C $Q_w$
2. formation of ice at 0 °C from the water of 0 °C $Q_L$
3. further cooling of ice of 0 °C to -20 °C $Q_i$

We have,

• Latent heat of fusion of ice, $L=3.4\times 10^5\ J.kg^{-1}$

• specific heat of water, $c_w=4186\ J.kg^{-1}.^{\circ}C^{-1}$

• specific heat of ice, $c_i=2000\ J.kg^{-1}.^{\circ}C^{-1}$

(a)

Now, total heat lost in the process:

$Q=Q_w+Q_L+Q_i$

$Q=m_w(c_w. \Delta T_w+L+c_i.\Delta T_i)$

where:

$\Delta T_i\ \&\ \Delta T_w$ = change in temperature of ice and water respectively.

$\Rightarrow Q=1.2(4186\times 20+3.4\times 10^5+2000\times 20)$

$Q=556464\ J$ is the total heat extracted during the process.

(b)

So, 556464 joule is the minimum electrical energy (by the law of energy conservation under no loss condition) required by refrigerator to carry out this process if it operates between the reservoirs at temperatures of 20.0 °C and -20.0 °C, because for a refrigerator to work in a continuous cycle it is impossible to transfer heat from a low temperature reservoir to a high temperature reservoir without consuming energy in the form of work. Here 556464 joule is the heat of the system to be eliminated.