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Sagot :
We can solve this question by using the principle of calorimetry.
The density of water is 1000 kg/m^3
According to the principle of calorimetry,
Heat lost by hot body = Heat gained by the cold body.
This principle is based on conservation of energy.
Let assume that the final temperature is t(f).
So,
[tex]\begin{gathered} Q=mc\Delta t \\ Where,\text{ } \\ Q=Heat \\ m=mass \\ c=Specific\text{ }heat\text{ }capacity \\ \Delta t=Change\text{ }in\text{ }temperature \end{gathered}[/tex]For two liter of water the final temperature is t(f) and initial temperature is 30°celcius,
So here the water will lose heat because it is at higher temperature,
[tex]\begin{gathered} Heat\text{ loose by the water is,} \\ Q=mc\Delta t=1000\times2\times c\times[t(f)-30] \end{gathered}[/tex]And for 1 liter of water the gain in heat is,
[tex]Q=mc\Delta t=1000\times1\times c\times[t(f)-20][/tex]Now,
[tex]\begin{gathered} Q\text{ }lose=Q\text{ }gain \\ -1000\times2\times c\times[t(f)-30]=1000\times1\times c\times[t(f)-20] \\ -2\times[t(f)-30]=[t(f)-20] \\ -2t(f)+60=t(f)-20 \\ -3t(f)=-20-60 \\ t(f)=\frac{80}{3}=26.667\mathring{\text{ }}celcius \end{gathered}[/tex]So the final temperature of the 3 liter water = 26.667 degree cus.
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