Question

A 31.1 g wafer of pure gold, initially at 69.3 °C, is submerged into 64.2 g of water at 27.8 °C, in an insulated container.


A 31.1 g wafer of pure gold, initially at 69.3 °C, is submerged into 64.2 g of water at 27.8 °C, in an insulated container. What the final temperature of both substances at thermal equilibrium?

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Answer #1

m(water) = 64.2 g

T(water) = 27.8 oC

C(water) = 4.184 J/goC

m(gold) = 31.1 g

T(gold) = 69.3 oC

C(gold) = 0.129 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by gold = heat gained by water

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

31.1*0.129*(69.3-T) = 64.2*4.184*(T-27.8)

4.0119*(69.3-T) = 268.6128*(T-27.8)

278.0247 - 4.0119*T = 268.6128*T - 7467.4358

T= 28.4107 oC

Answer: 28.4 oC

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