Question

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.

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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