Answer to Question #306298 in Physics for Phil

Question

How many calories of heat need to be extracted from 200.g of steam at 120.°C to change it into ice at --12.0°C ?

Answer 1

In order to cool down the steam to $100\degree C$ the following amount of heat should be extracted:

$$Q_1 =c_sm\Delta T$$

where $c_s = 0.48cal/g/\degree C$ is the specific heat capacity of steam, $m = 200g$ is the mass of steam, and $\Delta T = 120\degree C - 100\degree C = 20\degree C$. Thus, obtain:

$$Q_1 = 0.48\cdot 200\cdot 20 = 1920cal$$

In order to convert steem to water the following amount of heat should be extracted:

$$Q_2 = Lm$$

where $L = 540cal/g$ is the latent heat of vaporization of steam. Thus, obtain:

$$Q_2 = 540\cdot 200 = 108000cal$$

In order to cool down water from $100\degree C$ to $0\degree C$ the following amount of heat should be extracted:

$$Q_3 = c_wm\Delta T$$

where $c_w = 1.00cal/g/\degree C$ is the specific heat capacity of water, and $\Delta T = 100\degree C$. Thus, obtain:

$$Q_3 = 1\cdot 200\cdot 100 = 20000cal$$

In order co convert water to ice the following amount of heat should be extracted:

$$Q_4 = \lambda m$$

where $\lambda = 79.7cal/kg$ is the enthalpy of fusion of water. Thus, obtain:

$$Q_4 = 79.7\cdot 200 = 15940cal$$

In order to cool down the ice to $-12\degree C$ the following amount of heat should be extracted:

$$Q_5 = c_im\Delta T$$

where $c_i = 0.5cal/g/\degree C$ is the specific heat capacity of ice, and $\Delta T = 12\degree C$. Thus, obtain:

$$Q_5 = 0.5\cdot 200 = 100cal$$

The total heat is then:

$$Q = Q_1 + Q_2+Q_3+Q_4+Q_5 = 145960cal\approx 1.46\times 10^5cal$$

Answer. $1.46\times 10^5cal$