It takes much longer to boil off a certain quantitty of water than to bring it to its boiling pointfrom room temp.
Asked by lovemaan5500 | 26th Nov, 2017, 12:37: PM
This is due to latent heat of vapourisation of steam. Latent Heat of vapouraisation of steam is the heat energy required for
water to change its liquid state at boiling point 100°C to vapour state at 100°C. For water this energy is 22.5 x 105 J/kg.
For example if 1 kg of water is boiled from room teperature, say 25ºC to 100ºC, heat energy required is 1× 4200 ×(75) = 3.15 × 105 Joules.
For the same 1 kg water, heat energy required to undergo the change of state at boiling point 100 ºC, from liquid state at 100°C to vapour state 100°C is 22.5 × 105 Joules.
Hence if heat energy is supplied at same rate, it takes longer time for the water to get boiled off
Answered by | 27th Nov, 2017, 11:14: AM
- A piece of ice of mass 40g is added to 200g of water at 50 degree celcius
- a calorimetry of mass 50g contains 200g of water at 30degree celcius . Calculate the final temperature of the mixture when 40g of ice at -10degree celcius is added to it.
- find result of mixing 10g of ice at-10c of water at 10c
- Is heat a conserved quantity? Give reason in suport of your answer?
- The temperature of 170g of water at 50degree celsius is lowered to 5degree celcius by adding certain amount of ice to it.Find the mass of ice added . (Take ,specific heat capacity of water=4200J/kg0c and specific latent heat of ice =36000J/kg)
- Water falls from a height of 50m. Calculate the rise in the temperature of water when it strikes the bottom. G=10ms-2 C=4200J/Kg°C
- Detail solutions
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number