91Ó°ÊÓ

Given data:

Mass of metal is$125g$.

Temperature of the copper metal is$288°C$.

Mass of water is$85g$.

Temperature of the water is$12.0°C$.

Final temperature of water and metal is$24.0°C$.

The heat gained by water is equal to the heat lost by metal.

The equation to calculate the heat gained by water is,

$Heat=mass×temperaturechange×specificheat$

$\mathrm{q}={\mathrm{mC}}_{\mathrm{p}}\left(\mathrm{Δ}\mathrm{T}\right)$

Calculate the temperature change as follows:

The temperature changelocalid="1653054928045" $=\mathrm{Δ}\mathrm{T}$

localid="1653054930911" $=24°C-12°C$

localid="1653054933683" $=12°C$

Calculate the heat gained by water as follows:

localid="1653054937136" $\mathrm{q}=(85.0\mathrm{g})×\left(4.184\mathrm{J}/\mathrm{g}°\mathrm{C}\right)×\left(12°\mathrm{C}\right)$

localid="1651933377724" $=4267.68\mathrm{J}$

Calculate the temperature change as follows:

The temperature change$=\mathrm{Δ}\mathrm{T}$

$=288°\mathrm{C}-24°\mathrm{C}$

$=264°\mathrm{C}$

Calculate the heat gained by water as follows:

$q=\left(125g\right)×(specificheatofcopper)×(264°C)$

Calculate the specific heat of copper as follows:

Specific heat of copper$=\frac{q}{m×\mathrm{Δ}T}$

$=\frac{4267.68\mathrm{J}}{125×264}$

$=0.1293\mathrm{J}/\mathrm{g}°\mathrm{C}$

Therefore, specific heat of copper is

$0.1293\mathrm{J}/\mathrm{g}°\mathrm{C}$