Question

What mass of water at 23.0°C must be allowed to come to thermal equilibrium with a 1.80-kg cube of aluminum initially at 150°C to lower the temperature of the aluminum to 61.0°C? Assume any water turned to steam subsequently recondenses.
Your response is off by a multiple of ten. kg

My answer was 906.40 kg and I got it wrong

Answer 1

Using energy conservation:

Energy released by aluminum = energy gained by water

Q1 = Q2

ma*Ca*dT1 = mw*Cw*dT

ma = mass of aluminum = 1.80 kg

mw = mass of water = ?

dT1 = 150 - 61 = 89 C

dT2 = 61 - 23 = 38 C

Ca = heat capacity of aluminum = 900 J/kg-C

Cw = heat capacity of water = 4186 J/kg-C

So,

mw = ma*Ca*dT1/(Cw*dT2)

mw = 1.80*900*89/(4186*38)

mw = 0.906 kg = 906 gm

Please Upvote.

See that your units of final answer are wrong. Either use 0.906 kg OR 906 gm