Question

An electric heater of power 600 W raises the
temperature of 4.0 kg of a liquid from 10.0 °C to
15.0 °C in 100 s. Calculate: (i) the specific heat capacity of
liquid.​

Answer 1

Answer:

600 J/s * 100 s = 60,000 J

60,000 J / (4.19 J / cal) = 14,320 cal

14,320 cal / (5 degC * 4000 gm) = .72 cal / gm deg C