The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 7.40 g of condensation forms on a glass containing water and 210 g of ice, how many grams will melt as a result? Assume no other heat transfer occurs. Use the heat of vaporization of water at 37°, as an approximation, in this problem.
To calculate the amount of ice that will melt due to the condensation, we need to determine the heat transfer involved.
1. First, let's calculate the heat transferred when the condensation forms. We can use the equation:
Q = m * ΔHvap
Where:
Q is the heat transferred
m is the mass of condensation
ΔHvap is the heat of vaporization of water
Given:
m = 7.40 g
ΔHvap = Heat of vaporization of water at 37°C (Assuming 37°C is the temperature of the water)
ΔHvap = 2.257 kJ/g (approximation)
Q = 7.40 g * 2.257 kJ/g
Q = 16.6978 kJ
2. Next, let's calculate the amount of ice that will melt using the heat transferred. The heat transferred to the ice will cause it to melt, and the amount of heat needed to melt the ice can be calculated using:
Q = m * ΔHf
Where:
Q is the heat transferred
m is the mass of ice to be melted
ΔHf is the heat of fusion of water
Given:
Q = 16.6978 kJ
ΔHf = Heat of fusion of water
ΔHf = 334 kJ/kg (assumption that mass is in kg, as the units will cancel out later)
Rearranging the equation:
m = Q / ΔHf
m = (16.6978 kJ) / (334 kJ/kg)
m = 0.05 kg (or 50 g)
So, 50 grams of ice will melt as a result of the condensation forming on the glass.
To solve this problem, we need to calculate the heat transferred from the condensation to the ice, which will result in the melting of the ice. We can use the equation:
Q = m × ΔH
where Q is the heat transferred, m is the mass of the substance, and ΔH is the heat of fusion or heat of vaporization.
First, let's calculate the heat transferred by the condensation:
Q_condensation = m_condensation × ΔH_vaporization
Given that the m_condensation is 7.40 g and the heat of vaporization of water at 37°C is 2.26 × 10^6 J/kg, we need to convert the mass from grams to kilograms:
m_condensation = 7.40 g = 7.40 × 10^(-3) kg
Now, let's calculate the heat transferred by the condensation:
Q_condensation = (7.40 × 10^(-3)) kg × (2.26 × 10^6) J/kg
Q_condensation = 16.716 J
Since we assume no other heat transfer occurs, this amount of heat will be transferred from the condensation to the ice, causing it to melt. To calculate the mass of ice that will melt, we can use the equation:
Q_melt = m_melt × ΔH_fusion
where Q_melt is the heat required to melt the ice, and ΔH_fusion is the heat of fusion of ice, which is 3.33 × 10^5 J/kg.
Now, let's calculate the mass of ice that will melt:
Q_melt = m_melt × ΔH_fusion
m_melt = Q_melt / ΔH_fusion
Since the heat required to melt the ice comes from the condensation, we have:
Q_condensation = Q_melt
Therefore, we can substitute the values:
16.716 J = m_melt × (3.33 × 10^5) J/kg
Solving for m_melt:
m_melt = 16.716 / (3.33 × 10^5) kg
m_melt = 5.005 × 10^(-5) kg
Finally, to convert the mass from kilograms to grams:
m_melt = 5.005 × 10^(-5) kg = 5.005 × 10^(-2) g
Therefore, approximately 5.005 × 10^(-2) grams of ice will melt as a result of the condensation.