Question

The same quantity of heat was given to different masses of three substances A, B and C. The temperature rise in each case is shown in the table. Calculate the specific heat capacities of A, B and C.​

Answer 1

Given, the same quantity of heat was given to different masses of three substances A, B and C. The temperature rise in each case is shown in the table below.

$\begin{tabular}{|c|c|c|c|}\cline{1-4}Material & Mass/kg & Heat\:given/J & Temp.rise/^{\circ}C\\\cline{1-4}A&1.0&2000&1.0\\B&2.0&2000&5.0\\C&0.5&2000&4.0\\\cline{1-4}\end{tabular}$

To find, the specific heat capacities of A, B and C.

So, in this question we will apply the concept of specific heat capacity. i.e., The heat required to raise the temperature of 1 unit mass by 1 degree (either Celsius or Kelvin). Formula for specific heat capacity is

$\dashrightarrow\:\:\:\:\:\:\:\:\:\:\:\:{\boxed{\bf{q=c\times m\times\triangle T}}}$

where, q = heat required to raise temperature by 1°C

            c = specific heat capacity

            m = mass

          ΔT = temperature change (in °C)

So coming back to the question. We can also write the above formula as, $\bf{c=\frac{q}{m\times\triangle T}}$

1. Calculating the specific heat capacity of material A,

Here, m = 1.0 kg, q = 2000J and  ΔT = 1.0°C

Putting in the formula,

$\implies\sf{c=\frac{2000J}{1kg\times{1}^{\circ} C}}$

$\implies\bf{c=2000J/{kg}^{\circ}\:C}$

∴ the specific heat of material A is 2000J/kg°C.

_____________________________

2. Similarly calculating the specific heat capacity of material B,

Here, m = 2.0 kg, q = 2000J and  ΔT = 5.0°C

$\implies\sf{c=\frac{2000J}{2kg\times{5}^{\circ} C}}$

$\implies\sf{c=\frac{2000J}{10{kg}^{\circ} C}}$

$\implies\bf{c=200J/{kg}^{\circ}\:C}$

∴ the specific heat of material B is 200J/kg°C.

_____________________________

3. Calculating the specific heat capacity of material C,

Here, m = 0.5 kg, q = 2000J and  ΔT = 4.0°C

$\implies\sf{c=\frac{2000J}{0.5kg\times{4}^{\circ} C}}$

$\implies\sf{c=\frac{2000J}{2{kg}^{\circ} C}}$

$\implies\bf{c=1000J/{kg}^{\circ}\:C}$

∴ the specific heat of material C is 1000J/kg°C.

_____________________________

Note - Kindly visit the website of this app to view the table :)!!