Question

Two samples of copper metal (20.0g and 100.0g) are heated using boliing water (T=100C). The 20.0g sample is heated for 120 minutes white the 100.0g sample is heated for 60 minutes. Assuming it takes only 30 minutes for the metals to reach thermal equilibrium with the water regardless of the amount of metal, what can be said about the final temperature and energy content (the heat) of the metal samples?

Answer 1

Mass of sample 1 (m₁) = 20 g

Mass of sample 2 (m₂) = 100 g

sample 1 is heated to temperature (T₁) = 120 min

sample 2 is heated to temperature (T₂) = 60 mintues

thermal equilibrium time (T) = 30 minutes

Thus, there is large heat transfer in sample 1.since it is heated for large time

Now, change in temperature, dT = Q/(m*C)

where, Q =heat

m = mass

c= constant= specific heat of copper

for sample 1 mass is less, q islarge in comparions to sample 2 and thus change in temperature(dT) will be large. But for sample 2 change in temperture is comparatively low