JAMB - Physics (2009 - No. 13)
An electric heater rated 220v, 1000 w is immersed into a bucket full of water, calculate the mass of water if the temperature changes from 30°c to 100°c and the current flows for 300seconds.
[specific heat capacity of water = 4200 jkg-1 K-1]
[specific heat capacity of water = 4200 jkg-1 K-1]
4.28 kg
42.86 kg
1.02 kg
7.14 kg
Explanation
The energy supplied by the heater = heater gained by the water
pt = Mc\(\Delta\)θ
1000 x 300 = m x 4200(100 - 30)
m = \(\frac{(1000 \times 300)}{4200 \times 70}\) = 1.02 kg
pt = Mc\(\Delta\)θ
1000 x 300 = m x 4200(100 - 30)
m = \(\frac{(1000 \times 300)}{4200 \times 70}\) = 1.02 kg
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