Question

In an experiment, 3.9 g of steam at 100.0 ^oC is pumped into an aluminum caloroimeter with a mass of 91 g. It contains 111.2 g of water at 15 ^oC. What is the equilibrium temperature of the system?

Answer 1

here,

the mass of steam , m1 = 3.9 g

mass of alumunium , m2 = 91 g

mass of water , m3 = 111.2 g

let the final equilibrium temperature be Tf

for isolated system

using conservation of heat energy

heat gained by alumunium and water = heat lost by steam

m2 * Ca * ( Tf - 15) + m3 * Cw * ( Tf - 15) = m1 * ( Lv + Cw * ( 100 - Tf))

91 * 0.9 * ( Tf - 15) + 111.2 * 4.186 * ( Tf - 15) = 3.9 * ( 2250 + 4.186 * ( 100 - Tf))

solving for Tf

Tf = 33 degree C

the final temperature is 33 degree C