Question

A hot plate takes 17 minutes 55 seconds to boil 200 milliliters of water. The hot plate is raised to provide 200 Watts and is 30% efficient. What was the original temperature of the water? The specific heat of capacity of water is 4.18 joules per gram per degree Celsius.

Answer 1

Power provided to hot plate = 200 W

Time taken to boil the water = 17 mins 55 seconds = 17*60+55 seconds = 1075 s

Total energy consumed by hot plate = Power * time in seconds = 200*1075 = 215000 J

Since the hot plate is 30% efficient, actual heat provided = (total energy consumed *30)/100

Actual heat provided by hot plate = (215000*30)/100 = 64500 J

Given volume of the water = 200 mL

Density of water is nearly equal to 1g/mL.

Considering the same density, mass of water = 200 g = 0.2 Kg

Specific heat capacity of water = 4.18 J/g/degree Celsius

We know, heat required to raise temperature by amount dT = specific heat capacity* mass*dT

Substituting the values, we have

64500 = 4.18 * 200* dT

dT = 64500/(4.18*200) = 77.15 degree Celsius

Since water started to boil, final temperature is 100 degree Celsius.

dT = 77.15 = Final Temperature-Original Temperature

Original Temperature = 100-77.15 = 22.85 degree Celsius