Question #16158

Suppose 1900 J of heat energy are put into a sample of liquid water at 21°C, and the final temperature of the water is 76°C. How many grams of water are present?

Expert's answer

Heat capacity for water is C= 4218 J/Kg*K (it's for delta T =1 and 1 kg ) we

have delta T=55 and x kg

For 1 kg and delta T = 55 you need 55*4218 J=231990

j

For x kg and delta T = 55 we have 1900 J

x=1*1900/231990=0.0082 kg

or 8.2 ml

have delta T=55 and x kg

For 1 kg and delta T = 55 you need 55*4218 J=231990

j

For x kg and delta T = 55 we have 1900 J

x=1*1900/231990=0.0082 kg

or 8.2 ml

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