A sample of brass, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at . Calculate the initial temperature of the brass sample. Be sure your answer is rounded to significant digits.

The initial temperature of the brass sample is 90.1°C

Note: The question is incomplete. A similar but complete question is given below :

A 52.9g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is put into a calorimeter (see sketch at right) that contains 100.0g of water. The temperature of the water starts off at 15.0°C. When the temperature of the water stops changing it's 18.4°C. The pressure remains constant at 1 atm. Calculate the initial temperature of the brass sample. Be sure your answer is rounded to 2 significant digits.

Explanation:

Assuming that the calorimeter is an isolated system and that no heat is lost from the calorimeter. The total heat in the system is the sum of the heat content of the brass and that of water

Total heat lost by the brass = heat gained by the water

The quantity of heat lost or gained, Q = mcΔT

Where m = mass of the substance, c = specific heat capacity of substance, ΔT = temperature change

Heat gained by water is positive while heat lost by brass is negative

mass of brass = 52.9 g, specific heat capacity of brass = 0.375·J·g−1°C−1, ΔT = (18.4 - t °C; where t is the initial temperature), mass of water = 100.0 g, specific heat capacity of water = 4.186 J/g°C, ΔT = = 18.4 - 15.0 = 3.4 °C

Heat lost by brass z= - [ 52.9 × 0.375 × (18.4 - t)] = -365.01 + 19.8375t

Heat gained by water = 100 × 4.186 × 3.4 = 1423.24

Equating heat lost by brass to heat gained by water

-365.01 + 19.8375t = 1423.24

19.8375t = 1423.24 + 365.01

19.8375t = 1788.25

t = 90.1° C

Therefore, the initial temperature of the brass sample is 90.1°C

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