Asked by BOB
Work of 730 J is done by stirring an insulated beaker containing 80 g of water.
What is the change in the temperature of the water?
Please show calculation
What is the change in the temperature of the water?
Please show calculation
Answers
Answered by
drwls
The work done becomes the same as heat addition, Q = 730J = 174.5 calories.
Q = C M *(delta T)
C = 1.0 cal/(g*degC)
M = 80 g
Solve for deltaT, the water temperature change, in degrees C
deltaT = 174.5 cal/[(1.0 cal/degC)*(80 g)] = 2.2 deg C
Q = C M *(delta T)
C = 1.0 cal/(g*degC)
M = 80 g
Solve for deltaT, the water temperature change, in degrees C
deltaT = 174.5 cal/[(1.0 cal/degC)*(80 g)] = 2.2 deg C
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