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A 5.00-g sample of aluminum pellets (specific heat capacity 0.89 J/°C·g) and a 10.00-g sample of iron pellets (specific heat capacity 0.45 J/°C·g) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 97.3 g of water at 22.0°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Answer :

The final temperature of the metal and water mixture, assuming no heat loss to the surroundings is 23.7 °C.

The heat expression is given as :

q = mc ΔT

q = heat

m = mass

c = specific heat

ΔT = change in temperature

for the metals Al and Fe :

q = 5 × 0.89 × ( 100 - T) + 10 × 0.45 (100 - T)

for the water is given as :

q = 97.3 × 4.184 ( T - 22 )

the heat lost by the Al and Fe = heat gained by the water

5 × 0.89 × ( 100 - T) + 10 × 0.45 (100 - T) = q = 97.3 × 4.184 ( T - 22 )

415.95 T = 9851

T = 23.7 °C.

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