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const2013 [10]
3 years ago
15

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.
Physics
1 answer:
liq [111]3 years ago
6 0

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}

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