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?
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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