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4. A student collected the following data: 49.739 g water undergoes a temperature change of+23.48°C Cp for water is 4.184 J/g-C. What is the heat absorbed by this water sample?
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Answer #1

Heat absorbed by water = mass of water x Cp x temperature change

= 49.739 g x 4.184 J/gC x 23.48 C

= 4886.38 J x 1kJ/1000 J

= 4.88 kJ

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