Question

Two ice cubes of masses 100 g and 150 g initially at -5°C are added to 500 g of water at 40°C in an insulated vessel. Given that the specific heat of ice is 0.5 cal/g°C, that of water is 1 cal/g°C, and the latent heat of fusion for ice is 80 cal/g, what is the final equilibrium temperature?

Answer 1

0°C

Answer 2

The problem involves heat transfer between water and ice in an insulated vessel, leading to thermal equilibrium. We need to determine the final equilibrium temperature.

Let $m_1 = 100$ g and $m_2 = 150$ g be the masses of the two ice cubes. The total mass of ice is $m_i = m_1 + m_2 = 100 + 150 = 250$ g. The initial temperature of the ice is $T_i = -5^\circ$C. The mass of water is $m_w = 500$ g, and its initial temperature is $T_w = 40^\circ$C. The specific heat of ice is $s_i = 0.5$ cal/g$^\circ$C. The specific heat of water is $s_w = 1$ cal/g$^\circ$C. The latent heat of fusion of ice is $L_f = 80$ cal/g.

According to the principle of calorimetry, the heat lost by the hot body (water) is equal to the heat gained by the cold body (ice). The final equilibrium temperature, $T_f$, will be between the initial temperatures of the substances. Since ice is present, $T_f$ can be 0°C (if some ice remains) or above 0°C (if all ice melts).

First, let's calculate the heat required to bring the ice from -5°C to 0°C. $Q_1 = m_i \times s_i \times (0 - T_i) = 250 \times 0.5 \times (0 - (-5)) = 250 \times 0.5 \times 5 = 625$ cal.

Next, let's calculate the heat required to melt all the ice at 0°C. $Q_2 = m_i \times L_f = 250 \times 80 = 20000$ cal.

The total heat required to bring the ice from -5°C to water at 0°C is $Q_{ice\_to\_water\_at\_0} = Q_1 + Q_2 = 625 + 20000 = 20625$ cal.

Now, let's calculate the maximum heat that can be released by the water if it cools down from 40°C to 0°C. $Q_{water\_to\_0} = m_w \times s_w \times (T_w - 0) = 500 \times 1 \times (40 - 0) = 500 \times 40 = 20000$ cal.

We compare the heat required by the ice to melt completely (20625 cal) with the maximum heat available from the water by cooling to 0°C (20000 cal). Since $Q_{water\_to\_0} < Q_{ice\_to\_water\_at\_0}$, the water does not have enough heat to melt all the ice. This means that the final equilibrium temperature will be 0°C, with some ice remaining unmelted.

At equilibrium, the system will be a mixture of water and ice at 0°C. The heat supplied by the water as it cools from 40°C to 0°C is 20000 cal. This heat is used to first raise the temperature of the ice from -5°C to 0°C, which requires 625 cal. The remaining heat available for melting the ice at 0°C is $Q_{available\_for\_melting} = Q_{water\_to\_0} - Q_1 = 20000 - 625 = 19375$ cal. The mass of ice that melts is $m_{melted} = \frac{Q_{available\_for\_melting}}{L_f} = \frac{19375}{80} = 242.1875$ g. The initial mass of ice was 250 g. The mass of ice remaining is $250 - 242.1875 = 7.8125$ g. The final state consists of 500 g of original water, 242.1875 g of melted ice (now water), and 7.8125 g of ice, all at 0°C. Thus, the final equilibrium temperature is 0°C.