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

Answer:

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

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

$Q_1=m_1c_1\times (T-T_1)$

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

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(100 g\times 0.130 J/g^oC\times (10^oC-T_1))=400\times 4.186\times (10^oC-0^oC)$

we get, $T_1$ = 1,298°C

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