If 133.2 kJ of heat are added to 5.80 × 102 g of water at 22.0°C, what is the final temperature of the water? Specific heat of water at 1.00 atm and 20.0°C is 4.186 kJ/kg·K.
Specific of water is given as 4.186 kJ/Kg.K which means it will take 1kJ of energy to heat up 1Kg of water to raise its temperature by 1 degrees (Kelvin or celcius).
In this type of questions we use the formula to calculate the energy which is:
Now we are given E=133.2 KJ so,
So change(rise) in temperature is 54.9 oC. Hence the final temperature Tf is:
So final temperature is about 77o C or more precisely 76.9oC.
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