Question

A small electric heater is used to heat \(200\,\text{g}\) of water. The time required to bring all this water from \(40^\circ\text{C}\) to \(100^\circ\text{C}\) is \(200\,\text{s}\). If specific heat of the water is \(4200\,\text{J kg}^{-1}\text{K}^{-1}\), then the power supplied by the heater is

• A. \(155\,\text{W}\)
• B. \(310\,\text{W}\)
• C. \(88\,\text{W}\)
• D. \(252\,\text{W}\)

Hint:

For heating without phase change, use \[ Q=mc\Delta T \] and \[ P=\frac{Q}{t}. \] Always convert mass from grams to kilograms before substitution.

Answer 1

The Correct Option is D

Step 1: Calculate heat energy required.
Q = mcΔT = 0.2 × 4200 × 60 = 50400 J.

Step 2: Compute power from energy and time.
P = Q/t = 50400/200 = 252 W.

Step 3: Final Answer:
252 W.