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Ch205 Lesson 5 Essay examples

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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 …show more content…

q = ms∆T q = ? * 4.18J/g˚C * 40˚C
Mass of sample is not given, so we cannot calculate the heat added to the water sample.
Therefore the mass of the sample is needed to perform this calculation.
Part 3:
Two samples of water are heated from 20ºC to 60ºC. One of the samples requires twice as much heat to bring about this temperature change as the other. How do the masses of the two water samples compare? Explain your reasoning.
Answer:
Given;
Temperature change (∆T) = 60 – 20 = 40˚C
The first sample requires heat q1 = qJ (say) …….1
Then the second sample required heat q2 = 2q1J = 2q J …….2
Mass of first sample = m1g
Mass of second sample = m2g
Specific of water (s) = 4.18J/g˚C
By the equation: q = ms∆T q1 = m1 * 4.18J/g˚C * 40˚C for first sample q2 = m2 * 4.18J/g˚C * 40˚C for the second sample
Putting the value of q from equations (1) and (2)
2 * m1 * 4.18J/g˚C * 40˚C = m2 * 4.18J/g˚C we get:
2m1 = m2 m1 = ½ m2
Thus the mass of the first sample is one half the mass of the second sample.

6.30. Enthalpy * a. A 100.-g sample of water is placed in an insulated container and allowed to come to room temperature at 21ºC. To heat the water sample to 41ºC, how much heat must you add to it?
Answer:
100g water
Initial T = 21˚C
Final T = 41˚C
∆H = ?
From the equation: q = s * m * ∆T
∆T = tf – ti = 41˚C - 21˚C = +20˚C
Therefore, q = 4.18J/g˚C * 100g * (+20˚C) = 8.36 kJ

* b. Consider the hypothetical

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