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

`2365.71\ \text{J/kg}^(\circ)\text{C}`

Step-by-step 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=(VIt)/(m\Delta T)\\\Rightarrow c=(230* 1.8* 2* 60)/(0.75* (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}`