Question

An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

Answer 1

Step 1: Given data

1. Total energy rate supplied by the heater is $Q=100watt.$
2. Rate of Work done by the system is $W=75watt.$

Step 2: First law of thermodynamics

1. The first law of thermodynamics states that whenever work is transferred into heat or conversely heat into work, the quantity of work is mechanically equivalent to the quantity of heat.
2. The mathematical form of the first law is, $Q=W+U$, where, Q is the heat absorbed by the system, U is the increase in internal energy and W is the external work done by the system.

Step 3: Calculation of increasing internal energy

As we know, $Q=W+U$

or,

$$\begin{gathered} Q=W+U \\ orU=Q-W=100-75=25 \\ orU=25watt. \end{gathered}$$

Therefore, the internal energy will increase at a rate $25joule/sec.$