Question

1cc of water becomes 1681cc of steam when boiled at a pressure of 105Nm−2. The increasing internal energy of the system is
(Latent heat of steam is 540 cal g−1, 1 cal=4.2J)
(A). $300cal$
(B). $500cal$
(C). $225cal$
(D). $600cal$

Answer 1

- Hint: We have to use the first law of thermodynamics in order to calculate the internal energy. We have to find the total heat energy and the thermodynamic work. From the first law of thermodynamics the total heat will be the sum of internal energy and the thermodynamic work. Therefore by taking the difference of the total heat and thermodynamic work we can find the internal energy.

Formula used

$\Delta U = Q - dW$(First Law of Thermodynamics) ,where $Q$ is the heat energy, $U$ denotes the internal energy of the system and W denotes the thermodynamic work done
$Q = mL$, where $Q$ is the heat energy, $m$is the mass and $L$ is the specific latent heat.
$dW = pdV$ W denotes the thermodynamic work done $p$denotes the pressure and $dV$shows the change in volume.

Complete step-by-step solution -
Using the first law of thermodynamics we can say that internal energy difference can be calculated by taking the difference of heat energy and thermodynamic work. We have to calculate the heat energy.
$\Delta U = Q - dW$
Here the state of water has been changed from water to steam So we use latent heat of vaporization., which is a measure of the heat energy (Q) per mass (m) released or absorbed during a phase change.(Here water to steam.) Therefore,
$Q = mL$.
 A force acting on a system can heat it up (or cool it down), by working on it. Thermodynamic work can be calculated by taking the product of pressure and change in volume.
$dW = pdV$
Substituting the heat energy (Q) and thermodynamic work ($dW$) in first law of thermodynamics we get,
$\Delta U = (m \times L) - pdV$
Using$m = 1g$,$L = $$L = 540 \times 4.2J/g$, $p = {10^5}N/{m^2}$, $dV = (1681 - 1) \times {10^{ - 6}}{m^3}$
$\Delta U = (1 \times 2268) - ({10^5} \times 1680 \times {10^{ - 6}})$
$\Delta U = 2268 - 168$
$\Delta U = 2100J$
$\Delta U = (\dfrac{{2100}}{{4.2}})cal = 500cal$
Thus the correct option is B

Note: There are many cases of thermodynamics. Some are; Adiabatic process : In an adiabatic process there is no heat transfer into or out of the system. I.e. $Q = 0$ Therefore $\Delta U = - dW$ Isochoric Process: An isochoric process is one in which there is no change in volume I.e. $dW = 0$Therefore $\Delta U = Q$. Cyclic Process: The final state is the same as the initial state, so the total internal energy change must be zero. Therefore ,$Q = dW$.