Question

1. A 1000-kg block of ice melts in an arctic lake. The ice is at 0° C. The lake is initially at 10° C and cools to 5° C after the ice melts.
(a) What is the mass of the water in the lake?
(b) What is the heat lost by the lake due to the block of ice melting?

Answer 1

The heat that the lake losses is used to melt the ice (since it is already at melting temperature), and heat the water from ice from zero degrees to equilibrium temperature.

Hence, the balance equation is $c_w m_w (T^L_i - T) = \lambda m_i + c_i m_i (T - T_0)$, where

$T^L_i = 10^{\circ}C$ is the initial temperature of the lake, $T = 5^{\circ} C$ is the equilibrium temperature (of both lake water and ice water after melting), $T_0 = 0^{\circ} C$ is the melting point of the ice, $\lambda = 334 \frac{k J}{kg}$ is the latent heat of melting of ice, $c_i = 2100 \frac{J}{kg \cdot C}$ is the specific heat capacity of ice.

From the last equation, the mass of the water is $m_w = \frac{\lambda_i m_i + c_i m_i (T-T_0)}{c_w (T^L_i - T)} \approx 15905.3 kg$.

The heat lost by the lake is the quantity either in the left, or the right hand side of the balance equation, and is equal to $3.34 \cdot 10^8 J$.