WAEC - Physics (2005 - No. 19)
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]
1100Jkg-1K-1
320Jkg-1K-1
200Jkg-1K-1
176Jkg-1K-1
Wyjaśnienie
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