Answer to Question #201811 in General Chemistry for Candace Vessey

Question #201811

Mark is measuring the specific heat of a new type of coolant he first warms 75.0 g of this coolant from 23.5*C to 45.0*C then he puts the warm coolant into a calorimeter and cools it back to 23.5 *C the calorimeter calculations show the sample coolant shed 1370 calories of energy. What is the specific heat of the coolant?


1
Expert's answer
2021-06-02T06:30:54-0400

Q= 1370 cal

ΔT = 45.0-23.5=21.5

m=75.0 g

Q = mCΔT

"1370=75.0 \\times C \\times 21.5 \\\\\n\n1370 = 1612.5 \\times C \\\\\n\nC = \\frac{1370}{1612.5} = 0.849 \\;cal\/g \\; C"

The specific heat of the coolant is 0.849 cal/gC


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