Question

A19.0g sample of brass, which has a specific heat capacity of 0.375·J*g^−1°C−, is dropped into an insulated container containing 300.0g of water at 20.0°C and a constant pressure of 1atm. The initial temperature of the brass is 81.7°C.

1. Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has significant digits.

Answer 1

The final temperature is 20.3 °C

Step-by-step explanation:

Considering that:-

Heat gain by water = Heat lost by brass

Thus,

`m_(water)* C_(water)* (T_f-T_i)=-m_(brass)* C_(brass)* (T_f-T_i)`

Where, negative sign signifies heat loss

Or,

`m_(water)* C_(water)* (T_f-T_i)=m_(brass)* C_(brass)* (T_i-T_f)`

For water:

Mass = 300.0 g

Initial temperature = 20.0 °C

Specific heat of water = 4.184 J/g°C

For brass:

Mass = 19.0 g

Initial temperature = 81.7 °C

Specific heat of water = 0.375 J/g°C

So,

`300.0* 4.184* (T_f-20.0)=19.0* 0.375* (81.7-T_f)`

`1255.2T_f-25104=582.1125-7.125T_f`

`1262.325T_f=25686.1125`

`T_f = 20.3\ ^0C`

Hence, the final temperature is 20.3 °C