Question

What is the resultant temperature when 100g of steam at 100°C is passed through 500 g of ice at −20°C. The specific heat

of ice is 0.5 cal / g°C.

Answer 1

Q = mcΔT

Q = mL

Q(absorbed by ice) = Q(released by stem)

$Q_1 + Q_2 = Q_3 + Q_4 \\ m_{ice}L_f + m_{ice}c_i(T_f – T_i) = m_{steam}L_v + m_{steam}c_w(T_f -T_i) \\ L_f = 79.72 \; cal/g \\ L_v = 540 \; cal/g \\ c_w = 1 \; cal/gC$

$500 \times 79.72 + 500 \times 0.5 (T_f - (-20)) = 100 \times 540 + 100 \times 1 (T_f -100) \\ 39860 + 250T_f + 5000 = 54000 + 100T_f -10000 \\ 150T_f = -860 \\ T_f = -5.73 \; °C$