A 1-kg copper block at the initial temperature of 527 degrees C has been placed

Question

A 1-kg copper block at the initial temperature of 527 degrees C has been placed inside a perfectly insulated vessel containing 10 kg of water at the initial temperature of 27degrees C. The final temperature of the copper block is measured after the system has reached equilibrium. Assume the specific heat of copper equal to 0.38 kJ kg-1K-1.

(a) Calculate the final temperature of the copper block.
(b) Calculate the change in internal energy of the copper block between the initial and final states.

please show all working

Explanation / Answer

It is a very simple question which uses just one concept ,i.e., loss of heat by copper = gain of heat by water

So, let the final temperature of the system be X which lies between 27K and 527K.

therefore, mwcwA(delta Tw) = mccc(delta Tc)

10*1*(X-27) = 1*0.38*(527-X)

10X - 270 = 200.26 - 0.38X

10.38X = 470.26

X = 45.30 K

(b) Change in Internal Energy of Copper block = mc(delta T)

= 1*0.38*(527 - 45.3)

= 183 KJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.