Question

540 calories of heat convert 1 cubic centimetre of water at \[{100^o}\;C\;\] into 1671 cubic centimetre of steam at \[{100^o}\;C\;\] at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly:
A. 540 Cal
B. 40 Cal
C. zero Cal
D. 500 Cal

Answer 1

Hint:
This question can be solved by the concept of first law of thermodynamics and latent heat. First you need to identify the process involved in the conversion of water into steam and then apply the first law of thermodynamics.

Formula used:

\[\vartriangle Q = \vartriangle U + W\]

Complete step by step solution:

By looking at the question carefully, we see that water is at \[{100^o}\;C\] and also the ice is at \[{100^o}\;C\] temperature. Hence, no change in temperature. So, the process is isothermal because only change in states happens. Let’s discuss work done in an isothermal process.
An isothermal process is a process where the temperature of the system remains constant.
According to the first law of thermodynamics, the change in internal energy is the function only of temperature.
Hence, \[\vartriangle U = f\left( T \right)\] where T is the temperature, \[\vartriangle U\] is change in internal energy.
If temperature is constant then \[\vartriangle U = 0\].
Now, \[\vartriangle Q = \vartriangle U + W\] here \[\vartriangle Q\]is the net heat transfer and W is network done.
\[ \Rightarrow \vartriangle Q = W\]
For an isothermal process, \[W = \smallint PdV\] here P is pressure and V is volume. We know that$\;1{\text{ atmosphere pressure}} = {10^5}{\text{pascal}}$
Then, $W = p\Delta V$$ = {10^5} \times (1671 - 1) \times {10^{ - 6}} = {10^{ - 1}} \times 1670 = 167{\text{joule}}$
Now we know that ${\text{4}}{\text{.2joule = 1calories}}$.
Therefore, we can say that $167{\text{joule = (167/4}}{\text{.2)calories}}$$ = 40{\text{ calories}}$(approx.)




Therefore, option (B) is the correct option.



Note:
One should take care of the units of heat according to the options. To solve the question in the right way one should know the difference between specific heat and the latent heat.