5.29) The formation of condensation on a cold glass of water will cause it to warm up faster that it would have otherwise. If 8.0 g of water condenses on a 100 g glass containing 300 g of water at 5.0°C, what will be the final temperature? Ignore the effect of the surroundings.
Note. Provide an solution/formula to this problem. Thanks!
heat into water and glass from condensing = 80.0 grams * heat of vaporization of water
that heat = (T-5) ( 100*specific heat of glass + 300*specific heat of water)
To solve this problem, we can use the concept of specific heat capacity and the heat transfer equation. The formula we can use is:
Q = mcΔT
Where:
Q = Heat transferred
m = Mass
c = Specific heat capacity
ΔT = Change in temperature
We need to find the final temperature, so we rearrange the formula:
ΔT = Q / mc
First, we need to calculate the heat transferred (Q) when 8.0 g of water condenses. The heat released during condensation can be calculated using the formula:
Q = mL
Where:
m = Mass of water condensed
L = Latent heat of vaporization of water (normally around 2260 J/g)
Q = (8.0 g) (2260 J/g)
Next, we need to find the specific heat capacity (c) for the glass and water mixture. Specific heat capacity indicates how much heat energy is required to raise the temperature of a substance by 1 degree Celsius. For a mixture of water and glass, we can use an average value of 4.18 J/g°C as the specific heat capacity.
c = (mass of glass + mass of water) x specific heat capacity
c = (100 g + 300 g) × 4.18 J/g°C
Now, we can substitute the values into the formula to find the change in temperature:
ΔT = Q / (mc)
ΔT = (8.0 g) (2260 J/g) / ((100 g + 300 g) × 4.18 J/g°C)
After calculating the value of ΔT, we can determine the final temperature by adding the change in temperature to the initial temperature of 5.0°C:
Final temperature = Initial temperature + ΔT
Note: In this calculation, we are assuming that there is no heat exchange with the surroundings.