You have two identical solid objects (same mass mand specific heat c).they differ only in that one is at a temperature th0= 450 k and the other at tc0=200 k. there is no phase transition between these two temperatures. you run a heat engine between themuntil they are both at the same final temperature. to do the maximum amount of work, what would be that final temperature?

final temperature is 300K

Explanation:

Given data

temperature TH0 = 450 K

temperature TC0 = 200 K

to find out

final temperature

solution

we consider TH id temperature of hot and TC is temperature of cold object

so

dQH = -mc dTH   ..........1

dQC = mc dTC      ............2

and for max work

dQH/ TH = dQC/TC

put 1 and 2  here

-mc dTH / TH = mc dTC/ TC

now integrate both side from 450 to T and 250 to T

=

we get ln 450/T = ln T/200

so Temp = √(450*200)

Temp = 300K

final temperature is 300K

i don’t know what the answer is i wish i

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000