The specific heat of ice is 2.100 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, the heat of vaporization of water at 100.0 °C is 2,256 kJ/kg, and the specific heat of steam is 2.020 kJ/kg °C. What is the final equilibrium temperature when 20.00 grams of ice at −15.0 °C is mixed with 5.000 grams of steam at 120.0 °C?

Question

Quizplus · Accepted Answer

Explanation:The heat lost by the steam will be equal to the heat gained by the ice. The heat lost by the steam is:$$Q_{lost} = m_{steam} 	imes c_{steam} 	imes (T_{steam} - T_{final})$$$$Q_{lost} = (5.000 g) 	imes (2.020 kJ/kg °C) 	imes (120.0 °C - T_{final})$$The heat gained by the ice is:$$Q_{gained} = m_{ice} 	imes c_{ice} 	imes (T_{final} - T_{ice})$$$$Q_{gained} = (20.00 g) 	imes (2.100 kJ/kg °C) 	imes (T_{final} + 15.0 °C)$$Setting the heat lost equal to the heat gained, we get:$$(5.000 g) 	imes (2.020 kJ/kg °C) 	imes (120.0 °C - T_{final}) = (20.00 g) 	imes (2.100 kJ/kg °C) 	imes (T_{final} + 15.0 °C)$$Solving for $T_{final}$, we get:$$T_{final} = 59.92 °C$$

The Specific Heat of Ice Is 2