Question

Two balls of gold one 100g and other 225g at 23*C are dropped into 300.0mL of...

Two balls of gold one 100g and other 225g at 23*C are dropped into 300.0mL of hot water at 80.0*C. The specific heat of gold is 0.131J/g*C. Calculate the final temperature of the mixture.

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Answer #1

m(gold) = 100 g + 225 g = 325 g
T(gold) = 23.0 oC
C(gold) = 0.131 J/goC

Density of water is 1 g/mL and volume is 300.0 mL.
So,
m(water) = 300.0 g
T(water) = 80.0 oC
C(water) = 4.184 J/goC
T = to be calculated


Let the final temperature be T oC
use:
heat lost by water = heat gained by gold
m(water)*C(water)*(T(water)-T) = m(gold)*C(gold)*(T-T(gold))
300.0*4.184*(80.0-T) = 325.0*0.131*(T-23.0)
1255.2*(80.0-T) = 42.575*(T-23.0)
100416 - 1255.2*T = 42.575*T - 979.225
T= 78.13 oC
Answer: 78.1 oC

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