Question

A 31.1g wafer of pure gold initially at 69.4 degrees C is submerged into 63.2g of...

A 31.1g wafer of pure gold initially at 69.4 degrees C is submerged into 63.2g of water at 27.1 degrees C in an insulated container.

What is the final temperature of both substances at thermal equilibrium?

Homework Answers

Answer #1

Since the container is insulated, no heat is lost to the surroundings.

Thus the equation for heat balance would be:

Heat lost by gold wafer = Heat gained by water

mass of gold = 31.1 g

Specific heat capacity of gold = 0.129 J/oC g

Initial temperature = 69.4 oC

Final temeprature = T

Mass of water = 63.2 g

specific heat capacity of water = 4.186 J/ oC g

Initial temperature = 27.1 oC

Final temperature = T

Thus 31.1 g x 0.129 J/ oC g x (69.4 - T) oC = 63.2 g x 4.186 J/ oC g (T - 27.1) oC

278.42 - 4.01 T = 264.56 T - 7169.44

T = 27.73 oC

Thus, the final temperature of both substances is 27.73 oC at thermal equilibrium.

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