If 1.10 g of steam at 100.0 degrees celsius condenses into 38.5 g of water, initially at 27.0 degrees celsius, in an insulated container, what is the final temperature of the entire water sample? Assume no loss of heat into the surroundings
heat lost by 100 C water + heat gained by cold water - heat lost by condensation =0
1.1*4.184*(Tf-100) +[38.5*4.184*(Tf-27)] - 1.1*2257 = 0
Solve for Tf.
You can do it another way if you wish.
First calculate how much the condensing steam will raise the T of cold water.
1.1*2257 = 2482.7 J released.
38.5*4.184*(Tf-27) = 2482.7
Tf = 42.41 THEN
heat lost by 1.1g steam @ 100C + heat gained by 41.41 C water = 0 and recalculate Tf EXCEPT, of course, Tinitial is 42.41 and not 27.0 C. Same answer is obtained.
Thank you! what is the 2257 from?
2257 is the heat of vaporization of water.
To find the final temperature of the entire water sample, we can use the principle of energy conservation. The heat lost by the steam as it condenses is equal to the heat gained by the water as it warms up. We can calculate the heat lost by the steam and the heat gained by the water using the equations:
q1 = m1 * c1 * ΔT1 (heat lost by the steam)
q2 = m2 * c2 * ΔT2 (heat gained by the water)
where:
q1 and q2 are the heats lost and gained, respectively
m1 and m2 are the masses of the steam and water, respectively
c1 and c2 are the specific heat capacities of steam and water, respectively
ΔT1 and ΔT2 are the temperature changes of steam and water, respectively
Let's calculate the heats:
First, calculate the heat lost by the steam:
m1 = 1.10 g
c1 = specific heat capacity of steam = 2.03 J/(g °C) (provided in the question)
ΔT1 = final temperature - initial temperature of the steam = final temperature - 100.0°C
q1 = m1 * c1 * ΔT1
Next, calculate the heat gained by the water:
m2 = 38.5 g
c2 = specific heat capacity of water = 4.18 J/(g °C) (a commonly used value)
ΔT2 = final temperature - initial temperature of the water = final temperature - 27.0°C
q2 = m2 * c2 * ΔT2
Since the system is insulated and assuming no heat loss to the surroundings, we can equate q1 and q2:
q1 = q2
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Now, we can solve for the final temperature by rearranging the equation:
final temperature = ((m1 * c1 * ΔT1) / (m2 * c2)) + initial temperature of the water
Plug in the given values and solve the equation to find the final temperature of the entire water sample.