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
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
1 answer
1 answer