Question

A 30.45−gstainless steel ball bearing at 103.2°C is placed in a constant-pressure calorimeter containing 121.9 g of water at 19.55°C.

If the specific heat of the ball bearing is 0.474 J / (g·°C),

calculate the final temperature of both the water and steel when they equilibrate. Assume the calorimeter to have negligible heat capacity.

Answer 1

mass of steel bar = 30.45 g

temperature = 103.2 oC

mass of water = 121.9 g

temperature = 19.55 oC

here

heat loss by iron = heat gain by water

(m Cp dT)iron = (m Cp dT)water

30.45 x 0.474 x (103.2 - Tf) = 121.9 x 4.184 x (Tf - 19.55)

14.433 x (103.2 - Tf) = 510.0 x (Tf - 19.55)

(103.2 - Tf) = 35.33 x (Tf - 19.55)

(103.2 - Tf) = 35.33 Tf - 690.8

Tf = 21.86

final temperature = 21.86 oC