Question

40 kilograms of ice at -20c is placed in a chamber with 7 kilograms of steam at 103c. What is the final temperature of the system and what form is it in

Answer 1

Given, mass of ice, m1 = 40 kg

Initial temperature of ice, Ti = -20°C

Mass of steam m2 = 7 Kg

Initial temperature of steam, Ts = 103°C

The heat capacity of ice, Ci = 2108 J/Kg

The heat capacity of steam, Cs = 1996 J/kg

The latent heat of fusion, Lf = 3.36 ×10^5 J/kg

Latent heat of vaporization, Lv = 2.26 × 10^6 J/kg

Let the equilibrium temperature is, T°C in water form then,

Heat released = heat absorbed

m2×1996×(103-100) + m2×2.26×10^6 + m2×4186×(100-T) = m1×2108×20 +m1×3.36×10^5 + m1×4186×T

T = 1.28°C