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If a student adds $\pu{100.00 g}$ of hot water at $\pu{50.00 °C}$ to $\pu{100.00 g}$ of cold water at $\pu{25.00 °C}$ and measures a heat change of $\pu{10707.37 J}$ for the hot water and $\pu{4778.86 J}$ for the cold water, how much heat was lost to the calorimeter?

My thinking is assuming no energy is lost to the calorimeter. The amount of energy lost by the hot water should be equal to that gained by the cold water. It seems to me that the difference between $\pu{10707.37 J}$ and $\pu{4778.86 J}$ is the amount lost to the calorimeter.

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    $\begingroup$ Given the numbers, what are your lines of thought to determine the temperature of the two waters mixed together if there were no heat dissipation to the calorimeter? Add these in first place to document your attempts to solve this question on your own to show where you struggle. $\endgroup$
    – Buttonwood
    Oct 16, 2021 at 19:44
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    $\begingroup$ Note that the question makes no sense, if you mix two volumes of water, how would you measure the exact heat dissipated by one volume and that gained by the other? That aside, it is clearly a conceptual problem. To clarify the question, perhaps you might post the source. $\endgroup$
    – Buck Thorn ♦
    Oct 17, 2021 at 6:35

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I believe there is something wrong with the numbers provided in the question. Your thinking is right, assuming no energy is lost to the calorimeter, the heat lost by the hot water is the same as the gained by the cold water. So, by using the following formula: $$Q=mc\Delta T$$ We know the following: $$m(H_2O_{hot})\times c(H_2O)\times \Delta T(H_2O_{hot})=m(H_2O_{cold})\times c(H_2O)\times \Delta T(H_2O_{cold})$$ Let the final temperature be $T_f$, $$\implies mc(50-T_f)=mc(T_f-25)$$ Over here the unit of temperature (degree Celcius) does not affect the reading since we are subtracting the values (and therefore the difference would be the same irrespective of whether we are using degree Celcius or Kelvin as our unit). Now solving this equation for $T_f$, we get: $$T_f=\frac{75}{2} ^\circ C=37.5^\circ C$$ Now if we use $Q=mc\Delta T$ to find the value of heat loss for the hot water, we will see that it is not consistent with the values you have provided. This is shown below: $$Q=mc\Delta T$$ $$\implies Q=100g\times 4.184J\cdot g^{-1}\times (50-37.5)^\circ C$$ $$\implies Q=5230J$$ As seen, the heat change for hot water is $5230J$ worth of energy whereas in the question you have mentioned that the heat change for hot water is $10707.37J$. I would request you to recheck the question. But yes, as mentioned earlier, your concept is right, in an ideal case the heat change for hot water is the same as heat change for cold water (the signs will be opposite since hot water losses heat whereas cold water gains heat) and if it is not an ideal case, i.e. the calorimeter takes some of the heat energy then the heat lost to the calorimeter plus the heat change for cold water is equal to the heat change of the hot water.

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    $\begingroup$ Please post comments to the OP under the original question. I agree that there is something "wrong" with the question that the OP needs to address (then again, the question makes no sense, if you mix two volumes of water, how would you measure the exact heat dissipated by one volume and that gained by the other?) $\endgroup$
    – Buck Thorn ♦
    Oct 17, 2021 at 6:33
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    $\begingroup$ @BuckThorn I mean to measure the heat (energy) you can use $Q=mc\Delta T$... What I was saying was that the numbers given in the question are wrong. Will keep in mind about the commenting feature. $\endgroup$
    – noob anomaly
    Oct 17, 2021 at 6:53

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