A piece of aluminum (specific heat 0.910kJ/kg0C) of mass 138g at 740C is dropped

Question

A piece of aluminum (specific heat 0.910kJ/kg0C) of mass 138g at 740C is dropped into a Styrofoam cup filled with 159ml water at 200C. What are the final temperatures of the water and the aluminum? Please use C instead of 0C in your answer. Hint 1: because the cup is well-insulated, you can assume that all the heat leaving the aluminum as it cools goes to heating the water. Hint 2: How are the final temperature of the aluminum related to the final temperature of the water? (Use 4.19kJ/kgoC for the specific heat of water.)

Explanation / Answer

The heat Qout from the metal = heat Qin to the water.

In both cases Q = m*c*deltaT.

For the Al = 0.138 kg * 0.910 kJ/kgC * (74 - Tf) = 9.29 - 0.12558 Tf.

For water = 0.159 kg * 4.19 kJ/kgC * (Tf - 20) = 0.66621 Tf - 13.3242

Setting these equal gives,

9.29 - 0.12558 Tf = 0.66621 Tf - 13.3242

(0.66621 + 0.12558) Tf = 9.29 + 13.3242

Tf = (9.29 + 13.3242)/(0.66621 + 0.12558)

= 28.56 deg C

ANSWER

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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