If an average person jogs, he produces \(14.5 \times10^3\) cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming \(1\) kg requires \(580 \times10^3\) cal for evaporation) is:
1. | \(0.25\) kg | 2. | \(0.50\) kg |
3. | \(0.025\) kg | 4. | \(0.20\) kg |
Three copper blocks of masses \(M_{1} , M_{2} \) and \(M_{3}\) kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at \(T_{1} , T_{2} , T_{3} \left(\right. T_{1} > T_{2} > T_{3} \left.\right)\). Assuming there is no heat loss to the surroundings, the equilibrium temperature \(T\) is:
(\(s\) is the specific heat of copper)
1. \(T = \dfrac{T_{1} + T_{2} + T_{3}}{3}\)
2. \(T = \dfrac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}\)
3. \(T = \dfrac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{3 \left(\right. M_{1} + M_{2} + M_{3} \left.\right)}\)
4. \(T = \dfrac{M_{1} T_{1} s + M_{2} T_{2} s + M_{3} T_{3} s}{M_{1} + M_{2} + M_{3}}\)
(a) | \(dU = 0 \) | (b) | \(dQ = 0\) |
(c) | \(dQ = dU \) | (d) | \(dQ = dW\) |
1. | (a), (b), (c) | 2. | (a), (d) |
3. | (b), (c), (d) | 4. | (a), (c), (d) |
Consider a cycle followed by an engine (figure).
1 to 2 is isothermal,
2 to 3 is adiabatic,
3 to 1 is adiabatic.
Such a process does not exist, because:
(a) | heat is completely converted to mechanical energy in such a process, which is not possible. |
(b) | In this process, mechanical energy is completely converted to heat, which is not possible. |
(c) | curves representing two adiabatic processes don’t intersect. |
(d) | curves representing an adiabatic process and an isothermal process don't intersect. |
Choose the correct alternatives:
1. | (a), (b) | 2. | (a), (c) |
3. | (b), (c) | 4. | (c), (d) |