Question

A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me temperature by the process of heat transfer from copper to water. Calculate the final temperature of the water. The molar heat capacity of copper is 24.5 J · K-1 ·mol-1 and that of water is 75.2 1 · K -1 -mol-1•

Answer 1

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C×$\frac{1mol}{63,546g}$×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C×$\frac{1mol}{18,02g}$×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK×$\frac{1mol}{63,546g}$×25,0g×(X-363K) = -75,2J/molK×$\frac{1mol}{18,02g}$×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

X = 295K

Final temperature is 295K

I hope it helps!