A blackbody and a real body of identical dimensions are heated to same temperature. If ratio of rates of radiation of the blackbody and the real body is $4 : 3$, then emissivity of the real body is equal to
The rate of radiation is given by $E = e\sigma A T^4$. For a blackbody, $e = 1$. Given $\frac{E_b}{E} = \frac{4}{3} ⇒ \frac{1}{e} = \frac{4}{3}$, hence emissivity $e = 0.75$.
Water equivalent of a metallic bar is $100\text{ g}$, the energy required to increase its temperature by $1^\circ\text{C}$ is
Energy required is $Q = w \cdot \Delta T$, where $w$ is the water equivalent. Thus, $Q = 100\text{ g} \times 1\text{ cal/g}^\circ\text{C} \times 1^\circ\text{C} = 100\text{ cal}$.
For $n$ mole of an ideal gas, the correct equation of $1^{\text{st}}$ law of thermodynamics corresponding to isobaric process will be (symbols have their usual meanings)
By first law, $Q = \Delta U + W$. In an isobaric process, the work done is $W = P\Delta V = nR\Delta T$. Therefore, both expressions are correct.
In ideal condition, the maximum efficiency that can be derived from a heat engine operating between $600\text{ K}$ reservoir and $200\text{ K}$ sink, is
Efficiency is given by $\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$. Substituting the given values: $\eta = 1 -\frac{200}{600} =\frac{2}{3} \approx 66.67%$.
The equation of state for \(14\text{ g}\) nitrogen gas at a pressure \(P\) and temperature \(T\), when occupying a volume \(V\) will be
The molecular mass of nitrogen gas \((\text{N}_2)\) is \(28\text{ g/mol}\). The number of moles is \(n = \frac{14}{28} = 0.5\). Thus, using \(PV = nRT\), we get \(PV = \frac{1}{2}RT\).
Emissive power is defined as the thermal energy emitted per unit area per unit time, so its unit is \(\text{J m}^{-2}\text{ s}^{-1}\) (or \(\text{W m}^{-2}\)).
Two rods one made of material A and other made of material B of same length and same cross-sectional area are joined together. If thermal conductivity of material A is \(K_1\) while that of material B is \(K_2\) and the free end of rod made of material A is maintained at \(T_1\) while that of the rod of material B is maintained at \(T_2\), then the temperature of junction is (Where \(T_1 > T_2\))
Under steady state, the rate of heat flow is the same through both rods: \(\frac{K_1 A(T_1 - T_j)}{L} = \frac{K_2 A(T_j - T_2)}{L}\). Solving for \(T_j\) gives \(T_j = \frac{K_1 T_1 + K_2 T_2}{K_1 + K_2}\).
Consider the following statements out of which one is labelled as assertion and other as reason.
Assertion: The internal energy of an ideal monoatomic gas enclosed in a container does not change when there is no change in temperature.
Reason: Internal energy of a gaseous system is path function.
Internal energy of an ideal gas depends only on its temperature, so the Assertion is true. However, internal energy is a state function, not a path function, so the Reason is false.