Question

8.0g of aluminum at 200?C and 21g of copper are dropped into 45cm3 of ethyl alcohol at 15?C . The temperature quickly comes to 27?C.

What was the initial temperature of the copper?

Answer 1

Hi,

For quantity of heat gained or lost by the sample in this problem, we will be using the formula

q=mcdeltaT

Where,

q= heat gained or heat lost by the sample

m= mass of the sample

c= Specific heat of sample which is a constant at a given temperature

deltaT= change in Temperature

Assuming that: specific heat of aluminium is 0.900 J/gm.K

specific heat of copper is 0.386 J/gm.K

specific heat of ethyl alcohol is 2.4 J/gm.K

Heat given by aluminium would be

Q_Al= 8gm X 0.900 J/gm.K X (473K-288K) = 1332 J

Heat given by copper would be

Q_Cu= 21gm X 0.386 J/gm.K X (T-288K) = 8.106T-2334.528 J

Mass of ethyl alcohol = Density of ethyl alcohol X volume of ethylalcohol

Mass of ethyl alcohol= 789 kg/m^3 X 0.000045 m^3 = 0.03551 kg= 35.51 gm

Heat gained by ethanol = 35.51 gm X 2.4 J/gm.K (300K-288K) = 1022.688 J

Now, Heat lost = heat gained

1332 + 8.106T-2334.528 = 1022.688

8.106T = 2025.216

T= 249.84 K or -23.15^oC

This shows even copper has gained the heat

Hope this solves your query