Question

A 50 gram piece of copper at 200°C is placed in 100 grams of water at 25°C. Assuming no loss of heat to the surroundings, determine the final temperature of the water and copper.

Answer 1

specific heat capacity of copper = 0.385 J/g.oC

specific heat capacity of water = 4.184 J/g.oC

Let us denote Cu by symbol 1 and water by symbol 2

m1 = 50.0 g

T1 = 200.0 oC

C1 = 0.385 J/goC

m2 = 100.0 g

T2 = 25.0 oC

C2 = 4.184 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat gained by 2 = heat lost by 1

m2*C2*(T-T2) = m1*C1*(T1-T)

100.0*4.184*(T-25.0) = 50.0*0.385*(200.0-T)

418.4*(T-25.0) = 19.25*(200.0-T)

418.4*T -10460 = 3850 - 19.25*T)

T= 32.7 oC

Answer: 32.7 oC