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A continuous-flow calorimeter is used in two separate experiments to find the specific heat of a liquid. In the first trial, supplying 60 W raises the liquid temperature by 10 K. In the second trial, the power is doubled to 120 W, but the same 10 K rise is maintained by tripling the flow rate. Assuming heat loss to surroundings is the same in both trials, what is the power lost to the surroundings?

1. 20 W
2. 30 W
3. 40 W
4. 120 W

Correct answer: 20 W

Solution

Let q = m_dot*c*dT (heat carried per second in trial 1) and L = power loss. From trial 1: 60 = q + L. From trial 2: 120 = 3q + L. Subtracting gives 60 = 2q, so q = 30 W, and L = 60 - 30 = 20 W (Wait - let me recheck: 60 = q+L and 120 = 3q+L; subtracting: 60 = 2q => q=30; L = 60-30 = 20 W). Power lost = 20 W.

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