# NS Physics Question

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#### The_Juggernaut

##### New Member
15+ Year Member
Can someone please explain why c is the correct answer. I have no idea how they came up with this solution, nor do I understand their explanation.

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For this problem, it's possible to solve it the long way using thermal equilibrium using the q=mc(deltaT) equation. But that is going to be long as tedious with the decimals. The easiest way to solve it is using logic. We know that the water is 30 degrees warmer than the wood. In this case, heat is transferred from the water to wood. The definition of the specific heat of something is the amount of energy it takes to cause its temperature to increase by 1 degree Celsius/Kelvin per gram.

First, note that the mass of both objects in this question is identical since the density of water is 1kg/L. Now we note that the specific heat of water is larger than wood. In other words, it takes more energy to heat water, conversely water will release more energy for every 1 degree drop in temperature. Putting it all together now, we know that in thermal equilibrium an equal amount of energy is lost from 1 object, as is gained by other. Given what we know about the specific heats, we know that for every 1 degree lost by the water, the wood would gain a significantly larger number of degrees. Specifically, the ratio is 6 as they show in their solution.

Now thinking logically about the answers. Since the difference in temperature between objects is 30 degrees, the midpoint would be 15 degrees. But since the specific heats are different, it wouldn't make sense for the two objects to change by the same number of degrees. From above we know that the wood must change by a larger number of degrees than the water. So the only answer that makes sense is one that is larger than 15 which is choice (C). This logical approach only works because the masses are equal. If the masses were not equal you would need to do math to account for this.

If we wanted to calculate the exact answer, you could plug everything into the equation m1c1(deltaT1) = -m2c2(deltaT2) or use their approach of noting the ratio between specific heats of 6. From the ratio, you can then account for the 30 degree difference by counting by 1 degree changes from either side.
Boat: 0 6 12 18 24 30
Water: 30 29 28 27 26 25
Here you see the true final temperature change is somewhere between 25 and 26 degrees. Note I assumed the boat started at 0 degrees to simplify things, it doesn't matter what the starting temperatures actually are.