A puddle holds 150 g of water. If 0.50 g of water evaporates from the surface, what is the approximate temperature change of the remaining water? (Lv = 540 cal/g)
To determine the approximate temperature change of the remaining water, we need to calculate the amount of heat energy lost due to the evaporation of 0.50 g of water.
1. Convert the evaporation loss from grams to calories:
0.50 g * 540 cal/g = 270 cal
2. Determine the heat energy lost in terms of temperature change. We can use the specific heat capacity of water, which is approximately 1 calorie/gram°C.
Q = m * c * ΔT
ΔT = Q / (m * c)
ΔT = 270 cal / (150 g * 1 cal/g°C)
ΔT ≈ 1.8°C
Therefore, the approximate temperature change of the remaining water is approximately 1.8°C.
To calculate the approximate temperature change of the remaining water, we need to use the equation:
ΔQ = m × Lv
Where:
ΔQ is the heat transferred
m is the mass of the substance
Lv is the latent heat of vaporization
First, let's convert the given values into the correct units. The latent heat of vaporization (Lv) is given as 540 cal/g, and the mass (m) of the evaporated water is 0.50 g.
Now, since we know that heat is conserved, the heat lost by the water that evaporates is equal to the heat gained by the remaining water. We can set up the equation as follows:
ΔQ(evaporated) = -ΔQ(remaining)
m(evaporated) × Lv = m(remaining) × ΔT
Substituting the given values, we have:
(0.50 g) × (540 cal/g) = (150 g - 0.50 g) × ΔT
Now, we can solve for ΔT, the approximate temperature change:
(0.50 g) × (540 cal/g) = (149.50 g) × ΔT
Notice that the grams on the left side cancel out, and we are left with cal = cal × ΔT.
Dividing both sides by cal, we get:
1 = 149.50 × ΔT
Simplifying the equation, we have:
ΔT ≈ 1 / 149.50
Calculating this approximate value, we find:
ΔT ≈ 0.0067 degrees Celsius
Therefore, the approximate temperature change of the remaining water is approximately 0.0067 degrees Celsius.
heat released=50*heatvaporizatioin
look up the heat of vaporization of water.
Then, set that heat to = 100*specificeheatwater*deltaTemp
solve for deltaTemp