The specific heat of ice is 2.10 kJ/kg °C, the heat of fusion for ice at 0°C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, and the heat of vaporization of water at 100 °C is 2,256 kJ/kg. What is the final equilibrium temperature when 30.0 grams of ice at −15.0 °C is mixed with 8.00 grams of steam at 100 °C?

Question

Quizplus · Accepted Answer

First, we need to determine how much heat is required to raise the temperature of the ice to 0°C and then to melt the ice at 0°C:

Heat required to raise the temperature of ice: q1 = m × cice × ΔT = 30.0 g × 2.10 kJ/kg °C × [0 - (-15.0)] °C = 9,450 J

Heat required to melt the ice: q2 = m × ΔHfusion = 30.0 g × 333.7 kJ/kg = 10,011 J

Next, we need to determine how much heat is required to raise the temperature of the water from 0°C to the final temperature and then to vaporize the water at that temperature:

Heat required to raise the temperature of water: q3 = m × cwater × ΔT = 38.0 g × 4.186 kJ/kg °C × (Tf - 0) °C

Heat required to vaporize the water: q4 = m × ΔHvaporization = 38.0 g × 2,256 kJ/kg

The total heat added to the system is the sum of these four quantities:

qtotal = q1 + q2 + q3 + q4 = 9,450 J + 10,011 J + 38.0 g × 4.186 kJ/kg °C × (Tf - 0) °C + 38.0 g × 2,256 kJ/kg

Simplifying and solving for Tf:

Tf = (9,450 J + 10,011 J + 159.1 Tf + 85,728 J)/ (38.0 g × 4.186 kJ/kg °C + 38.0 g × 2,256 kJ/kg)

Tf = 65.6 °C

Therefore, the final equilibrium temperature is 65.6 °C. Answer choice (A) is the correct choice.

The Specific Heat of Ice Is 2