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?
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?
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a) To calculate the heat given up by the water to get cooled to the melting point, we need to use the equation:
Q = m * c * ΔT
Where:
Q = Heat energy
m = Mass
c = Specific heat capacity
ΔT = Change in temperature
For the water, the mass is 15 g and the change in temperature is (0°C - 55°C) = -55°C. The specific heat capacity of water is approximately 4.186 J/g°C.
So, using the equation above:
Q = 15 g * 4.186 J/g°C * -55°C
Q = -3439.65 J or approximately -3440 J
Therefore, the water would give up approximately -3440 J of heat to cool to the melting point.
b) To calculate the heat required to get the ice to the melting point, we use the same equation as before:
Q = m * c * ΔT
For the ice, the mass is 20 g and the change in temperature is (0°C - (-10°C)) = 10°C. The specific heat capacity of ice is approximately 2.09 J/g°C.
So:
Q = 20 g * 2.09 J/g°C * 10°C
Q = 418 J
Therefore, it would take approximately 418 J of heat to get the ice to the melting point.
c) Yes, some of the ice would melt because the water is at a higher temperature than the ice.
d) To calculate the amount of ice that would melt, we need to consider the heat of fusion, which is the amount of heat required to convert a substance from a solid to a liquid at its melting point.
The heat of fusion for water is approximately 334 J/g. So, for every gram of ice that melts, 334 J of heat are absorbed.
To find the amount of ice that would melt, we calculate the total heat given up by the water and divide it by the heat of fusion.
Total heat given up by water = Q from part a) + Q from part b)
= -3440 J + 418 J
= -3022 J
Amount of ice that would melt = (Total heat given up by water) / (Heat of fusion)
= (-3022 J) / (334 J/g)
≈ 9.05 g
Therefore, approximately 9.05 g of ice would melt.