An inventor develops a stationary cycling device by which an individual,while pedaling,can convert all of the energy expended into heat for warming water.What minimum power must be generated if 300 g water (enough for 1 cup of coffee) is to be heated in 12 min from 18°C to 94°C? (1 cal = 4.186 J,the specific heat of water is 4 186 J/kg⋅°C)

Question

An inventor develops a stationary cycling device by which an individual,while pedaling,can convert all of the energy expended into heat for warming water.What minimum power must be generated if 300 g water (enough for 1 cup of coffee)is to be heated in 12 min from 18°C to 94°C? (1 cal = 4.186 J,the specific heat of water is 4 186 J/kg⋅°C)
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Quizplus · Accepted Answer

The energy required to heat the water is calculated using the formula: Q = mcΔT. Here, m = 0.3 kg, c = 4186 J/kg°C, and ΔT = 94°C - 18°C = 76°C. Q = 0.3 * 4186 * 76 = 95,812.8 J. Power is energy divided by time, so P = 95,812.8 J / (12 * 60 s) = 133.07 W, which is approximately 130 W.

An Inventor Develops a Stationary Cycling Device by Which an Individual,while