Question

A student used apparatus as shown above. The beaker held 750 g of a liquid. The current from the power supply was 1.8 A and the potential difference of the supply was 230 V. The temperature rose from 12 °C to 40 °C over a period of 2 minutes.
Calculate the specific heat capacity of the liquid.

Answer 1

Answer:

$2365.71\ \text{J/kg}^{\circ}\text{C}$

Explanation:

V = Voltage = 230 V

I = Current = 1.8 A

$\Delta T$ = Temperature change = $(40-12)^{\circ}\text{C}$

t = Time taken = 2 minutes

m = Mass of liquid = 750 g

c = Specific heat capacity of the liquid

Energy required to heat the water is equal to the heat released due to current passing

$mc\Delta T=VIt\\\Rightarrow c=\dfrac{VIt}{m\Delta T}\\\Rightarrow c=\dfrac{230\times 1.8\times 2\times 60}{0.75\times (40-12)}\\\Rightarrow c=2365.71\ \text{J/kg}^{\circ}\text{C}$

The specific heat capacity of the liquid is $2365.71\ \text{J/kg}^{\circ}\text{C}$