Question

Please show all steps and reasoning to question.

A piece of copper with heat capacity 50 cal/C° is removed from an oven at 95°C and placed into a container with a 40-g sample of water at 20°C. The two substances are allowed to come into thermal equilibrium. (Note: Ignore thermal interactions with the air and the container itself.)

A. Consider the heat transfer needed to change the temperature of the water sample by 1C°. • How many calories are transferred? Explain. • What will be the temperature change (in C°) of the piece of copper? Show all work

B. Determine (i) the equilibrium temperature attained by the water and copper, as well as (ii) the total heat transferred. (Note: As before, “equilibrium” is considered to have been attained when the temperatures of the water and copper are within 0.1C° of each other.)

Answer 1

A.
Heat gained by water by raising 1 deg is, Q1 = mCT
m is the mass of water, m = 40 g
C is the specific heat of water = 1 cal/g ^oC
Q1 = 40 x 1 x 1 = 40 cal.

Heat lost by copper, Q2 = 50 x T
Initial temperature = 95 deg
Final temperature = Tf
Q2 = 50 x (95 - Tf)
Q1 = Q2
40 = 50 x (95-Tf)
95 - Tf = 0.8 degrees

B.
Consider that at temperature T, thermal equilibrium is attained.
Heat lost by copper = Heat gained by water
50 x (95 - T) = 40 x 1 x (T - 20)
475 - 5T = 4T - 80
9T = 555
T = 61.67 degrees.