A 2.0 kW heater is placed in 1.5 kg of water at 20 °C. The specific heat capacity of water is 4200 J kg−1 K−1 and no energy is lost. How long does the water take to reach 100 °C?
- Temperature change ΔT = 100 − 20 = 80 K.
- Energy needed: Q = mcΔT = 1.5 × 4200 × 80 = 504 000 J.
- time = energy ÷ power = 504 000 ÷ 2000 = 252 s.
Answer: About 250 s (a little over 4 minutes).