A heater coil connected to 220V supply, has a resistor of 150ohm. How long will this coil take to heat 1 kg of water from 20° to 60°C, assuming that all heat is taken up by water? (Given: heat capacity of water = 4186J/kg°C)
Answers
Answer:
Heat required to heat 1kg (=1000g) of water from 20°C to 60°C (rise of temperature ∆T = 60°C - 20°C = 40°C)= mass of water in grams × specific heat of water (1 calorie per gram of water per degree Centigrade rise of temperature) × rise of temperature in degree Celsius= 1000×1×40 = 40,000 calories
Heat generated per second by a heating coil of 150 ohm plugged into a 220 volt supply= V²/R Joule = (V²/R)/4.1840 Joule per calorie =(220²/4.184× 150 = 77.12 calorie
4.1840 Joule/calorie is the mechanical equivalent of heat ie it takes 4.2 Joule of work to generate 1 calorie if heat.
Time taken to generate 40,000 calorie of heat= 40,000/77.12 = 518. 7 sec= 8 minute 38.7 second.
It would take 8 m 38.7 s to heat 1kg of water from 20°C to 60°C.