# Heat Q flows spontaneously from a reservoir at 394 K

• copitlory8
In summary, the problem involves spontaneous heat flow from a reservoir at 394 K into a lower temperature reservoir T, resulting in 30% of the heat (Q) being unavailable for work when a Carnot engine operates between these two reservoirs at temperatures T and 248 K. The equation to solve for T is 0.3Q = (T - 248K) * ΔS, where ΔS is the change in entropy of the universe. Once T is solved for, it represents the temperature of the reservoir into which Q is flowing.
copitlory8
Heat Q flows spontaneously from a reservoir at 394 K into a reservoir that has a lower temerature T. Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature T and reservoir at 248 K. Find the temperature T.

First of all... I don't even understand the problem.

I know the change in entropy equals the heat (Q) divided by the temperature (T). I also know that the work unavailable (W) equals the temp of the coldest reservoir multiplied by the change of entropy of the universe.

I would appreciate any help! Thanks

.The temperature T is the temperature of the reservoir into which Q is flowing. The equation you need to solve is: 0.3Q = (T - 248K) * ΔS where ΔS is the change in entropy of the universe due to the flow of heat Q. Once you solve for T, you will have the temperature of the reservoir.

## 1. What is "heat Q"?

"Heat Q" refers to the transfer of thermal energy from one object to another. It is often represented by the symbol Q.

## 2. What is a "reservoir" in this context?

In this context, a reservoir refers to a large source of thermal energy that is at a constant temperature, meaning it can release or absorb heat without changing its own temperature. An example of a reservoir could be a large body of water or a furnace.

## 3. Why does heat Q flow spontaneously from a reservoir at 394 K?

This is due to the second law of thermodynamics, which states that heat always flows from hotter objects to colder objects in a natural, irreversible way. Since the reservoir is at a higher temperature of 394 K, heat will naturally flow from it to objects at lower temperatures.

## 4. Can heat Q flow in the opposite direction?

Yes, it is possible for heat Q to flow in the opposite direction if work is done on the system. For example, if a heat pump is used to transfer heat from a colder area to a warmer area, heat Q will flow in the opposite direction.

## 5. How is the amount of heat Q calculated in this scenario?

The amount of heat Q transferred can be calculated using the equation Q = mcΔT, where m is the mass of the object, c is its specific heat capacity, and ΔT is the change in temperature. In this scenario, the change in temperature would be the difference between the reservoir temperature of 394 K and the temperature of the object receiving the heat.

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