a mixture of 250g of water and 200g of ice at 0deg C is kept in a calorimeter which has a water equivalent of 50g. if 200g of steam at 100deg C is passed through this mixture, calculate the final temperature and the mass of contents of the calorimeter. ans 572.22g
please explain me the steps.
thanks in advance :)
Let the final temperature of the mixture = 100ºC.
Heat gained by ice and iced water Q1 is
Q1 = m•λ + m•c•ΔT = 200•80 + 200•1•100 = 3600 cal.
Q2 = (m1•c +W) •(100º -0º) =(250•1 + 50) •100 = 30000 cal.
Q =Q1+Q2 = 36000 +30000 = 66000 cal.
If entire steam condensed the total heat given by it is
Q3 = m•L = 200•540= 108000 cal.
Now the total heat available is 108000 cal, but only 66000 cal are required. Therefore? all the steam will not get condensed.
Final temperature of the mixture is 100ºC.
Mass of steam condensed is 66000/540=122.2 g.
Mass of the contents is 250 +200 + 122.2 = 572.2 g
Why did u assume that final temperature is 100 why not 70 or 80
Why do you take 66000/540 (at last).
You should take 108000-66000=42000
So,
42000/540???
To solve this problem, we can use the principle of conservation of energy. We can assume that there is no heat lost to the surroundings, so the heat gained by the ice and water mixture is equal to the heat lost by the steam.
1. Calculate the heat gained by the ice and water mixture:
The heat gained by the ice and water mixture can be calculated using the formula:
Q = m * c * ∆t,
where Q is the heat gained, m is the mass of the mixture, c is the specific heat capacity, and ∆t is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C.
∆t = final temperature - initial temperature = final temperature - 0°C
The mass of the mixture is the sum of the mass of water and ice:
mixture mass = mass of water + mass of ice = 250g + 200g = 450g.
Heat gained by the ice and water mixture:
Q1 = 450g * 4.18 J/g°C * (∆t)
2. Calculate the heat lost by the steam:
The heat lost by the steam can be calculated using the formula:
Q = m * ∆H, where Q is the heat lost, m is the mass of steam, and ∆H is the enthalpy of vaporization.
The enthalpy of vaporization of water is approximately 2260 J/g.
Heat lost by the steam:
Q2 = 200g * 2260 J/g
3. Equate the heat gained and heat lost:
Since there is no heat lost to the surroundings, we can equate Q1 and Q2:
Q1 = Q2
450g * 4.18 J/g°C * (∆t) = 200g * 2260 J/g
4. Solve for the final temperature (∆t):
Simplifying the equation:
450 * 4.18 * (∆t) = 200 * 2260
1877.7 * (∆t) = 452000
∆t = 452000 / 1877.7
5. Calculate the final temperature:
The final temperature is the initial temperature (0°C) plus ∆t:
final temperature = 0 + ∆t
6. Calculate the mass of the contents of the calorimeter:
The mass of the contents of the calorimeter is the sum of the mass of the mixture and the mass of the steam:
mass of contents = mixture mass + mass of steam = 450g + 200g = 650g.
Therefore, the final temperature of the mixture is calculated by dividing 452,000 by 1877.7, and the mass of the contents of the calorimeter is 650g.