TS EAMCET · Physics · Thermal Properties of Matter
The time required to raise the temperature of 3 litre of water from \(0^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) by a heater operated under \(200 \mathrm{~V}\) having resistance of \(50 \Omega\) is [specific heat capacity of water is \(4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\) ] [density of water \(=1000 \mathrm{~kg} / \mathrm{m}^3\) ]
- A \(12 \mathrm{~min}\)
- B \(18 \mathrm{~min}\)
- C \(21 \mathrm{~min}\)
- D \(24 \mathrm{~min}\)
Answer & Solution
Correct Answer
(C) \(21 \mathrm{~min}\)
Step-by-step Solution
Detailed explanation
Heat absorbed by water \(\Delta \mathrm{Q}_1=\mathrm{mS} \Delta \mathrm{T}\) \(=(3 \times 4200 \times 80) \mathrm{J} \quad[\because \mathrm{m}=\mathrm{V} \rho]\) Also, heat generated by heater…
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