Question

A 150.0-g sample of a metal at 75.0⁰C is added to 150.0 g of H₂O at 15.0⁰C. The temperature of the water rises to 18.3⁰C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

Short answer

The specific heat of the metal will be $0.24\frac{J}{g°C}$

Step-by-step solution

Determine the amount of heat gain

From the given information we can write Heat gain by water=Heat loss by metal

Let us assume,

$$\begin{gathered} m=massoftheelement \\ {C}_p=Heatcapacity \\ ∆T=Changeintemperature \\ Q=Amountofheat \end{gathered}$$

The amount of heat gained by water (H₂O) is:

$$\begin{gathered} Q=m{C}_p∆T \\ =150g\times 4.18\frac{J}{g°C}\times \left(18.3-15\right)°C \\ =2069.1J \\ =2.07kJ \end{gathered}$$

Determine the specific heat of the metal

The amount of heat required to raise one-degree Celsius temperature of one gram of substance is known to be the Specific heat capacity.

As Heat gain by water=Heat loss by metal therefore, Heat loss by metal=2.07 kJ

Now we need the determine the specific heat of the metal.

$$\begin{gathered} Q=m{C}_p∆T \\ 2.07kJ=150g\times C_p\frac{J}{g°C}\times \left(75-18.3\right)°C \\ 2070J=150g\times C_p\frac{J}{g°C}\times 56.7°C \\ {C}_p=0.24\frac{J}{g°C} \end{gathered}$$

Therefore, the specific heat of the metal will be$0.24\frac{J}{g°C}$