An unknown metal of mass 192 g heated to a temperature of 100$$^{\circ}$$C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4$$^{\circ}$$C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5$$^{\circ}$$C. (Specific heat of brass is 394 J kg$$^{-1}$$K$$^{-1}$$)

We have an unknown metal whose specific heat we call $$c_{m}\,( \text{J kg}^{-1}\text{K}^{-1})$$.

First, every mass must be written in kilograms because the S.I. unit of specific heat is J kg$$^{-1}$$K$$^{-1}$$.

$$m_{m}=192\;\text{g}=0.192\;\text{kg}$$ (mass of metal)

$$m_{w}=240\;\text{g}=0.240\;\text{kg}$$ (mass of water)

$$m_{c}=128\;\text{g}=0.128\;\text{kg}$$ (mass of brass calorimeter)

The temperatures are

$$T_{m}^{\,(i)} = 100^{\circ}\text{C},\qquad T_{w}^{\,(i)} = 8.4^{\circ}\text{C},\qquad T_{f}=21.5^{\circ}\text{C}.$$

The specific heats given or known are

$$c_{w}=4186\;\text{J kg}^{-1}\text{K}^{-1}\;(\text{water}),\qquad c_{c}=394\;\text{J kg}^{-1}\text{K}^{-1}\;(\text{brass}).$$

Energy conservation in calorimetry states:

$$\text{Heat lost by hot body} = \text{Heat gained by all cooler bodies}.$$

So,

$$m_{m}\,c_{m}\,(T_{m}^{\,(i)}-T_{f}) \;=\; m_{w}\,c_{w}\,(T_{f}-T_{w}^{\,(i)}) \;+\; m_{c}\,c_{c}\,(T_{f}-T_{w}^{\,(i)}).$$

Now we substitute every known quantity step by step.

The temperature fall of the metal is

$$T_{m}^{\,(i)}-T_{f}=100-21.5=78.5\;\text{K}.$$

The common temperature rise of water and the calorimeter is

$$T_{f}-T_{w}^{\,(i)}=21.5-8.4=13.1\;\text{K}.$$

Heat lost by the metal:

$$Q_{m}=0.192\,c_{m}\times 78.5 =0.192 \times 78.5 \; c_{m} =15.072\,c_{m}\;(\text{J}).$$

Heat gained by water:

$$Q_{w}=0.240 \times 4186 \times 13.1.$$

First multiply $$4186\times 13.1=54836.6,$$ then multiply by $$0.240$$ giving

$$Q_{w}=0.240 \times 54836.6 = 13160.784\;\text{J}.$$

Heat gained by the brass calorimeter:

$$Q_{c}=0.128 \times 394 \times 13.1.$$

Compute $$394\times 13.1 = 5161.4,$$ then multiply by $$0.128$$ giving

$$Q_{c}=0.128 \times 5161.4 = 660.6592\;\text{J}.$$

Total heat gained:

$$Q_{w}+Q_{c}=13160.784 + 660.6592 = 13821.4432\;\text{J}.$$

Set heat lost equal to heat gained:

$$15.072\,c_{m}=13821.4432.$$

Solve for $$c_{m}$$ by dividing both sides by 15.072:

$$c_{m}= \dfrac{13821.4432}{15.072}\;\text{J kg}^{-1}\text{K}^{-1}.$$

Carrying out the division,

$$c_{m}\approx 916\;\text{J kg}^{-1}\text{K}^{-1}.$$

Hence, the correct answer is Option A.