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 206-g cube of copper that is initially at 76°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?

Answer 1

Suppose the final temperature is T.

Now Using energy conservation:

Heat released/absorbed by Copper and Aluminum and water and steel = 0

Q1 + Q2 + Q3 + Q4 = 0

m1*C1*dT1 + m2*C2*dT2 + m3*C3*dT3 = 0

Q1 = Heat released by copper from 76 C to 25 C = m1*C1*dT1

Q2 = Heat absorbed by Aluminum from 5 C to 23 C = m2*C2*dT2

Q3 = Heat released/absorbed by water from 25 C to 25 C = m3*C3*dT3

Q4 = Heat released/absorbed by steel from 25 C to 25 C = m4*C4*dT4

dT1 = Tf - Ti = 25 - 76 = -51 C

dT2 = Tf - Ti = 25 - 5 = 20 C

dT3 = 25 - 25 = 0 C

m1 = 206 gm = 0.206 kg = mass of copper

m2 = mass of Aluminum = ?

mass of steel and water doesn't matter, since dT3 = 0, So

C1 = Specific heat capacity of Copper = 387 J/kg-C

C2 = Specific heat capacity of Aluminum = 900 J/kg-C

So,

0.206*387*(-51) + m2*900*20 + m3*C3*0 + m4*C4*0 = 0

m2 = 0.206*387*51/(900*20)

m2 = 0.226 kg = 226 gm

m2 = mass of aluminum chunk required = 226 gm

Let me know if you've any query.