Awater heater is operated by a solar power. if the solar collector has an area of 6 m2, and the intensity delivered by sunlight is 550 w/m2, how long does it take to increase the temperature of 1000 kg of water from 200c to 60 0c?

50 h

Explanation:

The heat required (Q) to heat the water  can be calculated using the following expression.

Q = c × m × ΔT = (4.184 kJ/kg.°C) × 1000 kg × (200°C - 60°C) = 5.9 × 10⁵ kJ

where,

c: specific heat capacity

m: mass

ΔT: change in the temperature

The solar collector has an area of 6 m² and the intensity delvered by sunlight is 550 W/m². The power (P) submitted by the collector is:

6 m² × 550 W/m² = 3 × 10³ W

The energy collected is transformed into heat. We know that:

P = E/t = Q/t

where

t: time

t = Q/P = 5.9 × 10⁸ J / (3 × 10³ J/s) = 2 × 10⁵ s × (1 h / 3600 s) = 50 h

what type of charge does it have? negative