6.29. Thermal Interactions
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)
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
= - 20˚C
Heat transfer from hot to cold water is q.
q = ms∆T
= 200g * 4.18J/g˚C * -20˚C
= -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 a sample of cold water that has twice its mass. Predict the temperature change of each of the samples. Answer:
If mass of hot water = 100g
Then mass of cold water = 200g
From the equation:
q = ms∆T for hot water
(temperature change) ∆T= 16720J100g*4.18J/g℃ = 40˚C
The final temperature of hot water = 120˚C
Initial temperature of hot water = 80˚C
The temperature change for hot water:
120˚C - 80˚C = 40˚C
Similarly for cold water:
q = ms∆T
(temperature change) ∆T= 16720J200g*4.18J/g℃ = 20˚C
The temperature change for cold water:
40 – 20 = 20˚C
The temperature change (∆T) for hot water = 40˚C
The temperature change (∆T) for cold water = 20˚C
* g. You mix two samples of water, and one increases by 20ºC, while the other drops by 60ºC. Which of the samples has less mass? How do the masses of the two water samples compare? Answer:
After mixing the two samples of water one increases by 20˚C while the other has dropped 60˚C. The first sample which increased by 20˚C has less mass than the second sample. The masses of the two samples compare by the heat capacity because the heat capacity is directly proportional to the amount of substance and is defined as the heat required for raising the temperature of the substance at a constant temperature. The small amount of sample requires less heat to raise the temperature and vice-versa. * h. A 7-g sample of hot water is mixed with a 3-g sample of cold water. How do the temperature changes of the two water samples compare? Answer:
After mixing the hot and cold water the 3g sample has increased its temperature from initial while the 7g sample has decreased its temperature from initial.
A sample of water is heated from 10ºC to 50ºC. Can you calculate the amount of heat added to the water sample that caused this temperature change? If not, what information do you need to perform this calculation? Answer:...