Question

A 50.0 gram sample of a metal initially at a temperature of 200 degrees celsius is immersed in 400 gram of water at 20 degrees celsius. Find the specific heat of the metal in J/kg degrees celsius, if the equilibrium temperature is 23 degrees celsius. (specific heat of water is 4186 J/kg degrees celsius)

Answer 1

Answer:

Specific heat of metal, $C_{m} = 567.59 J/kg$

Given:

Mass of metal, M = 50.0 g = 0.05 kg

Initial temperature, T = $200^{\circ}$

Mass of water, M' = 400 g = 0.4 kg

Temperature for water, T' = $20^{\circ}$

Equilibrium temperature, $T_{eqm} = [tex]23^{\circ}$

specific heat of water, $C_{w} = 4186 J/kg ^{\circ}C$

Solution:

At equilibrium:

specific heat of metal = specific heat of water

Therefore,

$MC_{m}\Delta T = M'C_{w}\Delta T$

$MC_{m}(T - T_{eqm}) = M'C_{w}(T' - T_{eqm})$

$C_{m} = \frac{M'C_{w}(T_{eqm} - T')}{M(T - T_{eqm})}$

$C_{m} = \frac{0.4\times 4186(23^{\circ} - 20^{\circ})}{50.0\times 10^{-3}(200^{\circ} - 23^{\circ})}$

$C_{m} = 567.59 J/kg$