Ch205 Lesson 5
Assignment Chapter 6
Concept Explorations
6.29. Thermal Interactions
Part 1:
In an insulated container, you mix 200. g of water at 80ºC with 100. g of water at 20ºC. After mixing, the temperature of the water is 60ºC. * a. How much did the temperature of the hot water change? How much did the temperature of the cold water change? Compare the magnitudes (positive values) of these changes.
200g of water at 80°C = hot water
100g of water at 20˚C = cold water
After mixing the temperature is 60˚C (equilibrium T)
Answer:
The temperature of hot water changed:
60˚C - 80˚C = -20˚C
The temperature of cold water changed:
60˚C - 20˚C = 40˚C
The temperature change hot water to cold water is 20:40.
* b. During the mixing, how did the heat transfer occur: from hot water to cold, or from cold water to hot?
Answer:
The heat transferred from hot water to cold water is due to their potential (temperature) difference.
* c. What quantity of heat was transferred from one sample to the other?
Answer:
Here heat transfer from hot water to cold water.
Mass of hot water (m) = 200g
Initial temperature (T) = 80˚C
Finial temperature (T2) = 60˚C
Specific heat of water (s) = 4.18 J/g˚C
∆T=(60-80)℃
= - 20˚C
Heat transfer from hot to cold water is q. q = ms∆T = 200g * 4.18J/g˚C * -20˚C = -16720J = -16720 * 10-3kJ q = -16.72kJ (1J = 10-3kJ)
Therefore heat transfer from hot to cold water = 16.72kJ
* d. How does the quantity of heat transferred to or from the hot-water sample compare with the quantity of heat transferred to or from the cold-water sample?
Answer:
* e. Knowing these relative quantities of heat, why is the temperature change of the cold water greater than the magnitude of the temperature change of the hot water.
Answer:
Since the mass of cold water (100g) is less than the mass of hot water (200g) the temperature change of cold water is greater than hot water.
* f. A sample of hot water is mixed with
Concept Explorations
6.29. Thermal Interactions
Part 1:
In an insulated container, you mix 200. g of water at 80ºC with 100. g of water at 20ºC. After mixing, the temperature of the water is 60ºC. * a. How much did the temperature of the hot water change? How much did the temperature of the cold water change? Compare the magnitudes (positive values) of these changes.
200g of water at 80°C = hot water
100g of water at 20˚C = cold water
After mixing the temperature is 60˚C (equilibrium T)
Answer:
The temperature of hot water changed:
60˚C - 80˚C = -20˚C
The temperature of cold water changed:
60˚C - 20˚C = 40˚C
The temperature change hot water to cold water is 20:40.
* b. During the mixing, how did the heat transfer occur: from hot water to cold, or from cold water to hot?
Answer:
The heat transferred from hot water to cold water is due to their potential (temperature) difference.
* c. What quantity of heat was transferred from one sample to the other?
Answer:
Here heat transfer from hot water to cold water.
Mass of hot water (m) = 200g
Initial temperature (T) = 80˚C
Finial temperature (T2) = 60˚C
Specific heat of water (s) = 4.18 J/g˚C
∆T=(60-80)℃
= - 20˚C
Heat transfer from hot to cold water is q. q = ms∆T = 200g * 4.18J/g˚C * -20˚C = -16720J = -16720 * 10-3kJ q = -16.72kJ (1J = 10-3kJ)
Therefore heat transfer from hot to cold water = 16.72kJ
* d. How does the quantity of heat transferred to or from the hot-water sample compare with the quantity of heat transferred to or from the cold-water sample?
Answer:
* e. Knowing these relative quantities of heat, why is the temperature change of the cold water greater than the magnitude of the temperature change of the hot water.
Answer:
Since the mass of cold water (100g) is less than the mass of hot water (200g) the temperature change of cold water is greater than hot water.
* f. A sample of hot water is mixed with