A 31.7 g wafer of pure gold initially at 69.9°C is submerged into 63.0 g of water at 26.9 °C in an insulated container.
Part A
What is the final temperature of both substances at thermal equilibrium?
m(water) = 63.0 g
T(water) = 26.9 oC
C(water) = 4.184 J/goC
m(gold) = 31.7 g
T(gold) = 69.9 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))
31.7*0.129*(69.9-T) = 63.0*4.184*(T-26.9)
4.0893*(69.9-T) = 263.592*(T-26.9)
285.8421 - 4.0893*T = 263.592*T - 7090.6248
T= 27.56 oC
Answer: 27.56 oC
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