A 31.7 g wafer of pure gold initially at 69.9 °C is submerged into 84.0 g...

Question

Question

Answer 1

Answer #1

m(water) = 64.0 g

T(water) = 27.1 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) = 64.0*4.184*(T-27.1)

4.0893*(69.9-T) = 267.776*(T-27.1)

285.8421 - 4.0893*T = 267.776*T - 7256.7296

T= 27.7438 oC

Answer: 27.7 oC

Answer 2

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