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How much heat is required to heat up 68 g of water by 9 degrees Celsius You must give correct units with your answer?
Wiki User
∙ 14y ago
The specific heat capacity of water is 4186J/kgK First convert your temperature to Kelvins: 9+273.15=282.15K
And g to kg: 68g*(1kg/1000g)=0.068kg
Now multiply through:
0.068kg*282.15K*4186J/kgK
All units cancel except J to get: 80,313J
Simplify to 80.3kJ (kilo joules)
Wiki User
∙ 14y ago
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19
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