What is the final temperature of a 100g ice at -15°C which absorbs 2000 calories of heat energy?
Specific heat of ice: ci=0.50cal/g°C
Specific heat of steam: cs=0.50cal/g°C
Specific heat of water: cw=1.00cal/g°C
Latent heat of fusion of ice: Lf=80Cal/g
Latent heat of Vaporization of Water: Lv=540Cal/g
No way, please check for typos
Here is what happens for 1 gram
-15 to 0 ==== 0.50*15 = 7.5
melt ======= 80, sum = 87.5
0 to 100===== 100 * 1 = 100 sum = 187.5
boil ======== 540 * 1 = 540 sum = 727.5
heat steam ======== .5 ( T-100)
so for just 1 gram you are up to 727.5 + 0.5(T-100)
for 100 grams you needed 72,750 just to get to steam at 100 C
sure it is not ten grams or 20,000 calories ?
nope. It's really 100g and 2000 calories.
since there is no mass of water...
q=mcΔT
2000J=100(0.50)(Tf-Ti)
2000J=100(0.50)(Tf-(-15))
Tf=25°C / 298K
To find the final temperature of the ice, we need to consider the different stages of the heating process and calculate the amount of heat required for each stage.
First, we need to raise the temperature of the ice from -15°C to 0°C. The amount of heat required for this stage can be calculated using the formula:
Q = m * ci * ΔT
where:
Q is the heat energy absorbed,
m is the mass of the ice (100g),
ci is the specific heat of ice (0.50cal/g°C), and
ΔT is the change in temperature (0°C - (-15°C) = 15°C).
So, Q = 100g * 0.50cal/g°C * 15°C = 750 calories.
Next, we need to melt the ice at 0°C to water at 0°C. The amount of heat required for this stage is equal to the latent heat of fusion of ice (Lf), which is 80Cal/g. Since 1 Calorie is equal to 1000 calories, the latent heat of fusion is 80 * 1000 = 80,000 calories.
Now, we need to raise the temperature of the water from 0°C to the final temperature. The amount of heat required for this stage can be calculated using the formula:
Q = m * cw * ΔT
where:
Q is the heat energy absorbed,
m is the mass of the water (100g),
cw is the specific heat of water (1.00cal/g°C), and
ΔT is the change in temperature yet to be found.
Since we don't know the exact final temperature, we'll let ΔT represent the difference between the final temperature (Tf) and 0°C. Therefore, ΔT = Tf - 0°C = Tf.
So, Q = 100g * 1.00cal/g°C * Tf = 100Tf calories.
Finally, we need to consider the phase change from water at the final temperature to steam. The amount of heat required for this stage is equal to the latent heat of vaporization of water (Lv), which is 540Cal/g. Again, since 1 Calorie is equal to 1000 calories, the latent heat of vaporization is 540 * 1000 = 540,000 calories.
Now, let's sum up all the heat energy absorbed in the different stages:
Q_ice_temperature_change + Q_ice_to_water + Q_water_temperature_change + Q_water_to_steam = Total heat energy absorbed
750 calories + 80,000 calories + 100Tf calories + 540,000 calories = 2000 calories
The 100Tf term represents the heat required to raise the temperature of the water from 0°C to the final temperature Tf. Rearranging the equation, we get:
100Tf = 2000 calories - 750 calories - 80,000 calories - 540,000 calories
100Tf = -619,750 calories
Tf = -619,750 calories / 100
Tf = -6197.5°C
The final temperature is -6197.5°C.
It's important to note that this calculated value is unrealistic since it is well below the boiling and freezing points of water. The negative value indicates that the ice would not completely melt, and the remaining ice would still be present at a lower temperature.