Problem Statement
An electric heater of power 1000 W is used to heat 2 kg of water from 25°C to 100°C. How long does it take? Specific heat of water = 4200 J/(kg·K).
Given Information
- Power P = 1000 W
- m = 2 kg
- ΔT = 100 − 25 = 75°C = 75 K
- c_w = 4200 J/(kg·K)
Physical Concepts & Formulas
The heater supplies energy at rate P. The heat required equals mcΔT. Time = energy / power.
- $Q = mc\Delta T$
- $t = \dfrac{Q}{P}$
Step-by-Step Solution
- $Q = 2 \times 4200 \times 75 = 630000$ J
- $t = \dfrac{630000}{1000} = 630$ s
Worked Calculation
$Q = 2 \times 4200 \times 75 = 630000$ J; $t = \dfrac{630000}{1000} = 630$ s
Answer
t = 630 s = 10.5 minutes
Physical Interpretation
A 1000 W heater takes 10.5 minutes to heat 2 L of water from room temperature to boiling. This matches everyday experience with electric kettles, which are typically 1500–3000 W and boil water in 3–5 minutes.
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