Question

A coin dealer, offered a rare silver coin, suspected that it might be a counterfeit nickel copy. The dealer heated the coin, which weighed 14.0 g to 100°C in boiling water and then dropped the hot coin into 22.0 g of water at T = 16.5°C in an insulated coffee-cup, and measured the rise in temperature. If the coin was really made of silver, what would the final temperature of the water be (in °C)? (for nickel, s = 0.445 J/g°C; for silver, s = 0.233 J/g°C )

Answer 1

m(water) = 22.0 g

T(water) = 16.5 oC

C(water) = 4.184 J/goC

m(silver) = 14.0 g

T(silver) = 100.0 oC

C(silver) = 0.232 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by silver = heat gained by water

m(silver)*C(silver)*(T(silver)-T) = m(water)*C(water)*(T-T(water))

14.0*0.232*(100.0-T) = 22.0*4.184*(T-16.5)

3.248*(100.0-T) = 92.048*(T-16.5)

324.8 - 3.248*T = 92.048*T - 1518.792

T= 19.346 oC

Answer: 19.3 oC