Advertisements
Advertisements
प्रश्न
An electric immersion heater is rated 1250 W. Calculate the time in which it will heat 20 kg of water at 5°C to 65°C.
Advertisements
उत्तर
Power supplied by an electric heater (VI) = 1250 W
θ1 = Initial temperature = 5°C
m = Mass of water = 20 kg
θ2 = Final temperature = 65°C
c = specific heat capacity by heater = Heat gained by water ...(Law of calorimeter)
VIt = mc (θ2 - θ1) .....(Q = VI t)
1250 t = 20 × 4200 × 60 ....[(65°C - 5°C = 60°C)]
1250 t = 5040000
t = 40325 seconds
APPEARS IN
संबंधित प्रश्न
What do you mean by the following statement?
The specific heat capacity of copper is 0. 4 Jg-1 K-1?
How will rise in sea level affect population in coastal countries?
What is meant by specific heat capacity?
A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg °C. It contains 250 gm of liquid at 30°C having specific heat of 0.4 kcal/kg °C. If we drop a piece of ice of mass 10 g at 0°C, What will be the temperature of the mixture?
(b) 2000 J of heat energy is required to raise the temperature of 4 kg of a
metal by 3°c. Which expression gives the specific heat capacity of the metal?
Explain, why is water sprayed on roads in evening in hot summer?
How much heat energy is released when 5 g of water at 20° C changes to ice at 0° C?
[Specific heat capacity of water = 4.2 J g-1 ° C-1 Specific latent heat of fusion of ice = 336 J g-1]
An office room contains about 4000 moles of air. The change in the internal energy of this much air when it is cooled from 34° C to 19° C at a constant pressure of 1.0 atm is (Use `gamma_"air"` = 1.4 and Universal gas constant = 8.314 J / mol K) ____________.
