Question

What mass of water at 25.0°C must be allowed to come to thermal equilibrium with a 1.85kg cube of aluminum initially at 1.50 x 102 °C to lower the temperature of the aluminum to 65.0°C? Assume any water turned to steam subsequently recondenses.

Answer #1

m_{w} = mass of water = ?

c_{w} = specific heat of water = 4186 J/(kg C)

T_{wi} = initial temperature of water = 25 C

T_{wf} = final temperature of water = 65 C

m_{c} = mass of cube = 1.85 kg

c_{c} = specific heat of cube = 900 J/(kg C)

T_{ci} = initial temperature of cube = 150 C

T_{cf} = final temperature of cube = 65 C

using conservation of heat

Heat gained by water = heat lost by cube

m_{w} c_{w} (T_{wf} - T_{wi} ) =
m_{c} c_{c} (T_{ci} - T_{cf})

m_{w} (4186) (65 - 25) = (1.85) (900) (150 - 65)

m_{w} = 0.85 kg

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