Question

An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5°C. (Specific heat of brass is 394 J kg⁻¹ K⁻¹)
(A) 458 J kg⁻¹ K⁻¹
(B) 1232 J kg⁻¹ K⁻¹
(C) 916 J kg⁻¹ K⁻¹
(D) 654 J kg⁻¹ K⁻

Answer 1

Given:

Mass of unknown metal $m_{1} = 192$ g

Temperature of metal $T_{1} =$ 100° C

Mass of brass $m_{3} = 128$ g

Mass of water $m_{2} = 240$ g

Temperature of brass $T_{2} =$ 8.4°C

Equilibrium temperature $T_{o} =$ 21.5°C

To Find:

Specific heat of unknown metal,

We know, Net heat loss = Net heat gain

    $m_{1} S\Delta T = m_{2} \Delta T + m_{3} \Delta T$

$192 \times S\times 78.5 = 128 \times 394 \times 13.1 + 240 \times 4200 \times 13.1$

  $S = \frac{128 \times 394 \times 13.1 + 240 \times 4200 \times 13.1}{78.5 \times 192}$

  $S = 916$ $\frac{J}{kg K}$

Therefore, the specific heat of metal is 916 $\frac{J}{kg K}$