Question on: WAEC Physics - 2005
An immersion heater rated 400W, 220V is used to heat a liquid of mass 0.5kg. If the temperature of the liquid increases uniformly at the rate of 2.5oC per second, calculate the specific heat capacity of the liquid. [Assume no heat is lost]
A
1100Jkg-1K-1
B
320Jkg-1K-1
C
200Jkg-1K-1
D
176Jkg-1K-1
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Correct Option: B
pt = mc\(\theta\)
C = \(\frac{pt}{m \theta}\)
= \(\frac{400 \times 1}{0.5 \times 2.5}\)
= 320Jkg-1K-1
C = \(\frac{pt}{m \theta}\)
= \(\frac{400 \times 1}{0.5 \times 2.5}\)
= 320Jkg-1K-1
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