Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32 °C such that the final temperature is 5 °C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1°C-1, latent heat capacity of ice = 330 J g-1.
Heat energy is supplied at a constant rate to 100g of ice at 0 °C. The ice is converted into water at 0° C in 2 minutes. How much time will be required to raise the temperature of water from 0 °C to 20 °C? [Given: sp. heat capacity of water = 4.2 J g-1 °C-1, sp. latent heat of ice = 336 J g-1].
The specific heat capacity of water is :
What impact will climate changes have on the crops of food?
How will global warming disturb the ecological balance?
What is carbon tax?
Write the approximate values of the specific latent heat of fusion of ice.
_______ is defined as the amount of heat required to raise the temperature of 1kg of a substance by 1°C.
Define specific heat capacity.
Which of the substances P, Q, or R has the lowest specific heat? The temperature v/s time graph is shown ______.

