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If 4.35 KJ of

heat is added to 15.5 g of water at 5.00 Celsius what will be the result in state and temp of the water

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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.
User Xiao Hanyu
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