Question

7. If 250-g of copper at 98oc is poured into a 60-g of aluminium cup that has 200-g of water (initially the water and cup are at 20oc), what will be the common final temperature of the mixture when equilibrium is reached, assuming no heat flows to the surroundings? 

Answer 1

here,

mass of copper , m1 = 250 g

mass of alumunium , m2 = 60 g

mass of water , m3 = 200 g

let the final temperature of the mixture at equilibium be Tf

as there is no heat transferred to the surrounding

heat gained bu alumunium and water = heat lost by copper

m2 * Ca * ( Tf - 20) + m3 * Cw * ( Tf - 20) = m1 * Cc * ( 98 - 20)

60 * 0.9 * ( Tf - 20) + 200 * 4.186 * (Tf - 20) = 250 * 0.385 * ( 98 - Tf)

solving for Tf

Tf = 27.6 degree C

the final temperature of the mixture at equilibium is 27.6 degree C