Question

1 kg of water initially at a temperature 27°C is heated by a heater of power 1KW. If the lid is opened, heat is lost at a constant rate of 200J/S. Find the time required for water to attain a temperature of 80°C with the lid open. (Specific heat of water = 4.2kJ/ kg/k)

Answer 1

Amount of water = 1 kg

Net heat supplied = (1000 – 200) = 800 W

Change in temperature = ∆ T

= (353 – 300) = 53K

∴ ∆ T = 53K

Time required, t = H/R

where H is the required energy and R is the rate of heat transfer 

H = mC ∆T

= 1 × 4.2 × 10³ × 53

= 2.226 × 10⁵ J

R = 8 × 10² J/s

∴ t = \(\frac{2.226 \times 10^5}{8\times 10^2}\)

= 2.783 × 10²

∴ t ≈ 279 = 4 min 39 secs.