Question

A $125g$piece of metal is heated to $288°C$and dropped into $85.0g$of water at $12.0°C$. The metal and water come to the same temperature of $24°C$. What is the specific heat, in $\mathrm{J}/\mathrm{g}{}^{\circ }\mathrm{C}$, of the metal?

Short answer

The heat gained by water is $4267.68\mathrm{J}$.

Specific heat of copper is $0.1293\mathrm{J}/\mathrm{g}°\mathrm{C}$.

Step-by-step solution

Given Information

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$24.0°C$.

Calculation of heat gained by water 

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

The equation to calculate the heat gained by water is,

$Heat=mass\times temperaturechange\times specificheat$

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

Calculate the temperature change as follows:

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

localid="1653054930911" $=24°C-12°C$

localid="1653054933683" $=12°C$

Calculate the heat gained by water as follows:

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

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

Calculation of specific heat of copper.

Calculate the temperature change as follows:

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

$=288°\mathrm{C}-24°\mathrm{C}$

$=264°\mathrm{C}$

Calculate the heat gained by water as follows:

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

Calculate the specific heat of copper as follows:

Specific heat of copper$=\frac{q}{m\times \Delta T}$

$=\frac{4267.68\mathrm{J}}{125\times 264}$

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

Therefore, specific heat of copper is

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