2. Suppose you pour 15 g of hot water (at 55 oC) into a glass that contains 20 g of ice (at - 10 oC). Neglect the heat absorbed in the glass itself.
a. How much heat would the water give up to get cooled to the melting point?
b. How much heat would it take to get the ice to the melting point?
c. Would some of the ice melt?
d. How much would melt?
To answer these questions, we’ll need to use the equations for specific heat and heat of fusion.
a. To determine the heat the water gives up to get cooled to the melting point, we need to calculate the heat lost using the formula:
Q = mcΔT
Where Q is the heat lost, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.
Given:
Mass of water, m = 15 g
Specific heat of water, c = 4.18 J/g°C
Change in temperature, ΔT = (0°C - 55°C) = -55°C (water is being cooled)
Using the formula:
Q = (15 g) * (4.18 J/g°C) * (-55°C)
Q = -3425 J
Therefore, the water would give up 3425 J of heat to get cooled to the melting point.
b. To determine the heat required to get the ice to the melting point, we need to calculate the heat gained using the formula:
Q = mL
Where Q is the heat gained, m is the mass of the ice, and L is the heat of fusion for ice.
Given:
Mass of ice, m = 20 g
Heat of fusion for ice, L = 334 J/g
Using the formula:
Q = (20 g) * (334 J/g)
Q = 6680 J
Therefore, it would take 6680 J of heat to get the ice to the melting point.
c. Yes, some of the ice would melt. The heat from the hot water will transfer to the ice, causing it to increase in temperature until it reaches its melting point.
d. To determine how much ice would melt, we need to compare the amount of heat given up by the water (calculated in part a) with the amount of heat needed to get the ice to the melting point (calculated in part b).
The heat given up by the water is 3425 J and the heat needed to get the ice to the melting point is 6680 J.
Since the heat given up by the water (3425 J) is less than the heat needed to get the ice to the melting point (6680 J), not all of the ice will melt. Only the amount of ice that corresponds to the heat given up by the water will melt.
To calculate the amount of ice that melts, we can use the equation:
melted ice mass = Q / L
Where Q is the heat given up by the water and L is the heat of fusion for ice.
Using the equation:
melted ice mass = (3425 J) / (334 J/g)
melted ice mass = 10.26 g
Therefore, approximately 10.26 g of ice will melt.
To answer these questions, we can use the concept of heat transfer and the specific heat capacities of water and ice. Here's how we can approach each question:
a. To find out how much heat the water would give up to get cooled to the melting point, we need to know the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Step 1: Calculate the temperature change of the water from 55°C to 0°C.
ΔT = Final temperature - Initial temperature
ΔT = 0°C - 55°C = -55°C
Step 2: Calculate the heat lost by the water using the formula:
q = m * c * ΔT
where q is the heat energy in Joules, m is the mass of the water in grams, c is the specific heat capacity in J/g°C, and ΔT is the temperature change in °C.
Substituting the values:
q = 15 g * 4.18 J/g°C * -55°C
q = -3448.5 J
Therefore, the water would give up approximately 3448.5 Joules of heat to get cooled to the melting point.
b. To find out how much heat it would take to get the ice to the melting point, we need to know the specific heat capacity of ice. The specific heat capacity of ice is approximately 2.09 J/g°C.
Step 1: Calculate the temperature change of the ice from -10°C to 0°C.
ΔT = Final temperature - Initial temperature
ΔT = 0°C - (-10°C) = 10°C
Step 2: Calculate the heat gained by the ice using the formula:
q = m * c * ΔT
where q is the heat energy in Joules, m is the mass of the ice in grams, c is the specific heat capacity in J/g°C, and ΔT is the temperature change in °C.
Substituting the values:
q = 20 g * 2.09 J/g°C * 10°C
q = 418 J
Therefore, it would take approximately 418 Joules of heat to get the ice to the melting point.
c. Since the temperature of the water is higher than the melting point of ice, the water will release heat to the ice. This heat transfer will cause some of the ice to melt.
d. To find out how much ice would melt, we need to consider the heat lost by the water (calculated in part a) and the heat gained by the ice (calculated in part b). The heat lost by the water would be equal to the heat gained by the ice, as no heat is lost or gained in the process.
So, the amount of ice that would melt can be calculated by dividing the heat gained by the ice by the heat of fusion of water, which is approximately 334 J/g.
Amount of ice melted = Heat gained by the ice / Heat of fusion of water
Amount of ice melted = 418 J / 334 J/g
Amount of ice melted ≈ 1.25 g
Therefore, approximately 1.25 grams of ice would melt.