Question

A platinum ball with a mass of 100 g is removed from a furnace and dropped into 400 g of water at 0 ◦C. If the equilibrium temperature is 10◦C and the specific heat of platinum is 130 J/kg◦C, what was the temperature of the furnace?

Answer 1

1,298 °C was the temperature of the furnace.

Step-by-step explanation:

Heat lost by platinum will be equal to heat gained by the water

`-Q_1=Q_2`

Mass of platinum =
`m_1=100 g`

Specific heat capacity of platinum=
`c_1=130 J/kg^oC =0.130 J/g^oC`

Initial temperature of the iron =
`T_1=?`

Final temperature =
`T_2`=T = 10°C

`Q_1=m_1c_1* (T-T_1)`

Mass of water=
`m_2= 400 g`

Specific heat capacity of water=
`c_2=4.186 J/g^oC`

Initial temperature of the water =
`T_3=0^oC`

Final temperature of water =
`T_2`=T = 10°C

`Q_2=m_2c_2* (T-T_3)`

`-Q_1=Q_2`

`-(m_1c_1* (T-T_1))=m_2c_2* (T-T_3)`

On substituting all values:

`-(100 g* 0.130 J/g^oC* (10^oC-T_1))=400* 4.186* (10^oC-0^oC)`

we get,
`T_1` = 1,298°C

1,298 °C was the temperature of the furnace.