Question

A hot tub contains 1.00 x 106 grams of water. The heater for this hot tub...

A hot tub contains 1.00 x 106 grams of water. The heater for this hot tub can heat at a rate of 2500 kJ/min.

If the hot tub heater is on for 75.0 minutes, and the initial temperature of the water was 26 C, what will be the new temperature of the water? (The specific heat of water is 4.18 J/g C)

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Answer #1

Given weight of water = 10e6 g

Given power rating = 2500kJ / min = 2.5e6 J / min

We know by conservation of heat energy, that heat gained by a body H = mc

where H = energy supplied, m = mass of body , c = sp. heat of body , = change in temperature

Heat H = Power * time = 75 * 2.5e6 = 1.875e8 J
Change in temperature = H / mc =  1.875e8 J / (10e6 g * 4.18 J/g C) = 44.856 C

Initial temperature = 26 C

Hence, final temperature = 26 + 44.856 = 70.856 C

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