AP EAMCET · PHYSICS · Thermal Properties of Matter
A small electric heater is used to heat \(200 \mathrm{~g}\) of water. The time required to bring all this water from \(40^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) is \(200 \mathrm{~s}\). If specific heat of the water is \(4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\) then the power supplied by the heater is
- A \(155 \mathrm{~W}\)
- B \(310 \mathrm{~W}\)
- C \(88 W\)
- D \(252 \mathrm{~W}\)
Answer & Solution
Correct Answer
(D) \(252 \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
Given, \(t=200 \mathrm{~s}, m=200 \mathrm{~g}=0.2 \mathrm{~kg}\) \(\begin{aligned} s & =4200 \mathrm{jkg}^{-1} \mathrm{~K}^{-1} \\ \Delta T & =(100+273)-(40+273)=60 \mathrm{~K}\end{aligned}\) Heat absorbed by water = Energy supplied by heater \(\Rightarrow\) Power of heater…
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