Question

A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 180-g cube of copper that is initially at 75°C, and the other is a chunk of aluminum that is initially at 5.0°C. To the surprise of the student, the water reaches a final temperature of 25°C, precisely where it started. What is the mass of the aluminum chunk in grams?

Answer 1

Suppose mass of aluminum chunk is = Ma

Now using energy conservation:

Heat gained by water + steel container = Heat released by copper + aluminum

Heat gained or released due to temperature increase/decrease is given by = Q = m*C*dT

C = specific heat of material, & m = mass of object & dT = Change in temperature

So,

Q1 + Q2 = Q3 + Q4

Mw*Cw*dT1 + Ms*Cs*dT2 = Mc*Cc*dT3 + Ma*Ca*dT4

dT1 = dT2 = 25 - 25 = 0 C

So, Q1 = Q2 = 0

Q3 = -Q4

|Q3| = |Q4|

Mc*Cc*dT3 = Ma*Ca*dT4

dT3 = 75 - 25 = 50 C

dT4 = 25 - 5 = 20 C

Cc = 385 J/kg-C

Ca = 902 J/kg-C

Mc = 180 gm = 0.18 kg

Ma = Mc*Cc*dT3/(Ca*dT4)

Ma = 0.18*385*50/(902*20)

Ma = 0.192 kg = 192 gm

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