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3 years ago

If 4.35 KJ of

1 answer:

4 0

4.186 Joules equals 1 cal.
1 cal is the amount of energy needed to increase in 1ºC the temperature of 1 gram of water. If we add to 15.5 grams of water 4350 Joules, we are adding approximately 1039.2 cal. (4350÷4.186≈1039.2)
This energy though is being divided by all the gram in the water. 1039.2cal÷15.5g≈67.0cal/g
4.35kJ would increase the water's temperature in approximately 67ºC, putting it at 72ºC. The state would remain the same - liquid - though much closer to boil.

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Explanation:

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The final temperature is obtained by following relation at constant volume

$\displaystyle \frac{P}{P'}=\displaystyle \frac{T}{T'} \\ \displaystyle \frac{845 \textrm{ mmHg}}{1050 \textrm{ mmHg}} = \displaystyle \frac{328 \textrm{ K}}{T'} \\T' = 407.57 \textrm{ Kelvin}$

The final temperature is 407.57 K

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