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

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As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle your physics question. To find the rate at which the internal energy is increasing, we need to consider the energy conservation principle,...
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As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle your physics question.

To find the rate at which the internal energy is increasing, we need to consider the energy conservation principle, which states that the change in internal energy of a system is equal to the heat supplied to the system minus the work done by the system.

Given: Heat supplied (Q) = 100 W (Watts) Work done (W) = 75 J/s (Joules per second)

We know that power is the rate of doing work or the rate at which energy is transferred or converted. So, the rate at which work is done is the same as power.

Therefore, we have: Rate of work done (W)=Power (P)=75 J/sRate of work done (W)=Power (P)=75J/s

Now, let's calculate the rate at which internal energy is increasing using the formula:

Rate of change of internal energy=Heat supplied−Rate of work doneRate of change of internal energy=Heat supplied−Rate of work done

Rate of change of internal energy=100 W−75 J/sRate of change of internal energy=100W−75J/s

Rate of change of internal energy=25 J/sRate of change of internal energy=25J/s

So, the internal energy is increasing at a rate of 25 Joules per second. If you need further clarification or assistance with any other concepts, feel free to ask! And remember, UrbanPro is the best online coaching tuition platform to find experienced tutors like myself.

 
 
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