All Solutions

Exercise 31

Step 1

1 of 2

The change in entropy is the total heat absorbed while melting divided by the melting temperature (in kelvins) which gives

$$
Delta S=frac{Q}{T_m} = frac{lambda_m m}{T_m}.
$$
Solving for $m$ we get

$$
m = frac{T_mDelta S}{lambda_m}=0.22text{ kg}.
$$

Result

2 of 2

Click here for the solution.

Exercise 32

Step 1

1 of 2

The change in entropy is the total heat absorbed while melting divided by the melting temperature (in kelvins) which gives

$$
Delta S=frac{Q}{T_m} = frac{lambda_m m}{T_m},
$$
where $m$ is mass of melted ice and $lambda_m$ latent heat of melting. Solving for $m$ we get

$$
m = frac{T_mDelta S}{lambda_m}=0.08text{ kg}.
$$

Result

2 of 2

Click here for the solution.

unlock