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

Question

Quizplus · Accepted Answer

The correct answer is A because the final equilibrium temperature is 65.6 ^$\circ$C. The heat gained by the ice is equal to the heat lost by the steam. The heat gained by the ice is:Q = m * c * ΔT = (0.030 kg) * (2.10 kJ/kg ^$\circ$C) * (15 ^$\circ$C) = 0.945 kJThe heat lost by the steam is:Q = m * L = (0.008 kg) * (2,256 kJ/kg) = 18.048 kJThe total heat gained by the ice and the steam is:Q = 0.945 kJ + 18.048 kJ = 18.993 kJThe final equilibrium temperature is:T = Q / (m * c) = (18.993 kJ) / ((0.030 kg) * (4.186 kJ/kg ^$\circ$C)) = 65.6 ^$\circ$C

The Specific Heat of Ice Is 2 ∘^{\circ}∘ C,the Heat of Fusion for Ice at 0

The Specific Heat of Ice Is 2 $^{\circ}$ C,the Heat of Fusion for Ice at 0