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

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) = (M'C_(w)(T_(eqm) - T'))/(M(T - T_(eqm)))`

`C_(m) = (0.4* 4186(23^(\circ) - 20^(\circ)))/(50.0* 10^(-3)(200^(\circ) - 23^(\circ)))`

`C_(m) = 567.59 J/kg`