7. 50g of steam at 100 degree centigrade is passed into a mixture of 100g of ice and 200gm of water at 0 degree centigrade. Find the rise in temperature. (S.L. H. Of steam = 540 cal/g and Lf of ice = 80cal/g)
To find the rise in temperature, we need to calculate the heat transferred.
First, let's calculate the heat required to melt the ice:
Heat required = Mass of ice * Latent heat of fusion of ice
= 100g * 80 cal/g
= 8000 cal
Next, let's calculate the heat required to raise the temperature of the resulting water to the final temperature:
Heat required = Mass of water * specific heat capacity of water * temperature change
= 200g * 1 cal/g°C * (final temperature - 0°C)
= 200g * (final temperature - 0°C)
= 200g * (final temperature)
Now, let's calculate the heat released by the steam to condense into water:
Heat released = Mass of steam * latent heat of condensation
= 50g * 540 cal/g
= 27000 cal
Since heat is conserved in this process (no heat is lost or gained), we can set up an equation:
Heat released = Heat required to melt the ice + Heat required to raise the temperature + Heat released by the steam
27000 cal = 8000 cal + 200g * (final temperature) + 0
Now, let's solve for the final temperature:
200g * (final temperature) = 27000 cal - 8000 cal
200g * (final temperature) = 19000 cal
(final temperature) = 19000 cal / 200g
(final temperature) = 95°C
Therefore, the rise in temperature is 95°C.
To find the rise in temperature, we need to calculate the amount of heat transferred during the process.
First, let's calculate the heat required to convert the steam into water at 100 degrees Celsius.
Heat required = mass × specific latent heat of vaporization
= 50g × 540 cal/g
= 27000 cal
Now, the steam condenses into water, releasing 27000 calories of heat. This heat is used to raise the temperature of the ice and water mixture.
The heat released by the condensing steam is given by the formula:
Heat released = mass × specific heat capacity × change in temperature
Let's assume the rise in temperature is ΔT.
For the ice:
Heat released by 100g of ice = 100g × 80 cal/g × ΔT
For the water:
Heat released by 200g of water = 200g × 1 cal/g/°C × ΔT
The total heat released is equal to the heat required for the condensation of steam.
100g × 80 cal/g × ΔT + 200g × 1 cal/g/°C × ΔT = 27000 cal
Simplifying the equation:
8000 ΔT + 200 ΔT = 27000
8200 ΔT = 27000
ΔT = 27000 / 8200
ΔT ≈ 3.29 °C
Therefore, the rise in temperature is approximately 3.29 degrees Celsius.
I'm going to generalize.
mL (steam) released by cooling steam
provides heat for:
mc change T cold water, cooled steam and melted ice
mL (ice)
The mL's are straight forward
Watch mc change T terms. You have cooled steam going from 100 to Tf and the melted ice and cold water going from 0 to Tf.