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 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
Show 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.
Updated On: Jun 18, 2026
- \(155\,\text{W}\)
- \(310\,\text{W}\)
- \(88\,\text{W}\)
- \(252\,\text{W}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the given data.
Mass of water is \[ m=200\,\text{g} \] \[ m=0.2\,\text{kg} \] Initial temperature is \[ T_1=40^\circ\text{C} \] Final temperature is \[ T_2=100^\circ\text{C} \] So, change in temperature is \[ \Delta T=100-40 \] \[ \Delta T=60\,\text{K} \] Specific heat of water is \[ c=4200\,\text{J kg}^{-1}\text{K}^{-1} \] Time taken is \[ t=200\,\text{s} \]
Step 2: Calculate the heat supplied.
The heat required is \[ Q=mc\Delta T \] Substituting the values, \[ Q=0.2\times 4200\times 60 \] \[ Q=50400\,\text{J} \]
Step 3: Calculate the power supplied by the heater.
Power is given by \[ P=\frac{Q}{t} \] \[ P=\frac{50400}{200} \] \[ P=252\,\text{W} \]
Step 4: Final conclusion.
Therefore, the power supplied by the heater is \[ \boxed{252\,\text{W}} \]
Mass of water is \[ m=200\,\text{g} \] \[ m=0.2\,\text{kg} \] Initial temperature is \[ T_1=40^\circ\text{C} \] Final temperature is \[ T_2=100^\circ\text{C} \] So, change in temperature is \[ \Delta T=100-40 \] \[ \Delta T=60\,\text{K} \] Specific heat of water is \[ c=4200\,\text{J kg}^{-1}\text{K}^{-1} \] Time taken is \[ t=200\,\text{s} \]
Step 2: Calculate the heat supplied.
The heat required is \[ Q=mc\Delta T \] Substituting the values, \[ Q=0.2\times 4200\times 60 \] \[ Q=50400\,\text{J} \]
Step 3: Calculate the power supplied by the heater.
Power is given by \[ P=\frac{Q}{t} \] \[ P=\frac{50400}{200} \] \[ P=252\,\text{W} \]
Step 4: Final conclusion.
Therefore, the power supplied by the heater is \[ \boxed{252\,\text{W}} \]
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