physics problem about energy

Derek Lee

2010-04-06 5:19 am

An iron cube with length of each side 5 cm is heated from 50 degree celsius to
200 degree celsius at 1 atm
For iron, specific heat capacity = 0.46 J/g degree celsius and the volume
coefficient of thermal expansion is 3.6 x 10^-5 degree celsius^-1.
The mass of the cube is 1500g

a)How much work is done?
b)What is the change in internal energy?

回答 (1)

天同

2010-04-06 5:56 am

✔ 最佳答案

(a) Initial volume of cube = 5 x 5 x 5 = 125 cm^3
Change in volume = 125 x 3.6 x 10^-5 x (200-50) cm^3 = 0.675 cm^3 Hence, work done against atmospheric pressure in expansion
= (1.01 x10^5) x (0.675 x 10^-6) J = 0.0682 J

(b) Heat supplied to heat the iron cube
= 1500 x 0.46 x (200-50) J = 103 500 J

By First Law of Thermodynamics
Q = dU + W
where Q is the heat supplied
dU is the change in internal energy
W is the work done

Hence, dU = Q - W = (103500 - 0.0682) J = 103500 J
That is to say, the change done is negligibly small. Nearly all of the energy supplied goes to the increase in internal energy of the iron cube.

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