WBJEE · Physics · Thermal Properties of Matter
A scientist proposes a new temperature scale in which the ice point is \(25 \mathrm{X}(\mathrm{X}\) is the new unit of temperature) and the steam point is \(305 \mathrm{X}\). The specific heat capacity of water in this new scale is \(\left(\operatorname{in} J \mathrm{kg}^{-1} \mathrm{X}^{-1}\right)\)
- A \(4.2 \times 10^{3}\)
- B \(3.0 \times 10^{3}\)
- C \(1.2 \times 10^{3}\)
- D \(1.5 \times 10^{3}\)
Answer & Solution
Correct Answer
(D) \(1.5 \times 10^{3}\)
Step-by-step Solution
Detailed explanation
Given, \(305 \times-25 X=100^{\circ} \mathrm{C}\) (: \(\mathrm{X}\) is the new unit of temperature \[ (305-25) \mathrm{X}=100^{\circ} \mathrm{C} \Rightarrow 280 \mathrm{X}=100^{\circ} \mathrm{C} \] \(\therefore \quad 1^{\circ} \mathrm{C}=2.8 \mathrm{x}\) The specinc heat…
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