The correct answer is 42
\(Q = msΔT\)
\(\frac{dQ}{dt} = (\frac{dm }{dt})_{water}\)
\( SΔT = (\frac{dm}{dt})_{oil}C\)
\(⇒ 2 × 4.2 × 10³ × 40\)
\(= (\frac{dm}{dt})_{oil} × 8 × 10^6\)
\(⇒ (\frac{dm}{dt})_{oil} = \frac{8 × 4.2 × 10^4}{8 × 10^6} kg/minute\)
= 42 g/minute

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
From the second law of thermodynamics, two important results are derived where the conclusions are taken together to constitute Carnot’s theorem. It may be stated in the following forms.
