Q153 / 200 · WBJEE 2014 · Physics
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$1.2 \times 10^{3}$
- Correct answerD$1.5 \times 10^{3}$
Solution
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 capacity of water
$\begin{aligned}=4200 \frac{j 0 u l e}{k g \cdot^{\circ} C} &=4200 \times \frac{j \text { oule }}{k g \times 2.8 \times} \\ &=1500 J / \mathrm{kg}-\mathrm{X} \\ &=1.5 \times 10^{3} \mathrm{J} \mathrm{kg}^{-1} \mathrm{X}^{-1} \end{aligned}$
Hence, the correct answer is option: 'D'.