160g of steam at 100°C is added to 1000 g of ice at 0°C. What is the equilibrium temperature of the mixture? Assume the container has a negligible heat capacity.
Qloss+Qgain=0;[m*c(delta T)] + [m*c(delta T)] = 0
The Attempt at a Solution
i have the final answer which is 19.3 degree Celsius but i dunno how to get it., i used the formula of Qloss+Qgain=0;[m*c(delta T)] + [m*c(delta T)] = 0. but my final answer is not 19.3.., can you please help me? what formula will i need to use? do i need to consider the latent heat of vaporization and fusion?if so, what should be the formula?
this is my solution;
Qloss(steam) + Qgain (ice) = 0
m*C (ΔT) + m*C (ΔT) = 0
160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0
76.8 (t-100) + 500 (t-0) = 0
76.8 t - 7680 + 500 t = 0
576.8 t / 576.8 = 7680/576.8
t = 13.3 °C