1. How much heat must be added to 40 g of water at an initial temperature of 64°
ID: 1731182 • Letter: 1
Question
1. How much heat must be added to 40 g of water at an initial temperature of 64°C to do the following? (a) heat it to the boiling point, (b) to completely convert the 100°C water to steam? (Show your work)
2. Suppose that the pressure of an ideal gas mixture remains constant at 1900 Pa (1 Pa = 1 N/m2) and the temperature is increased from 200 K to 850 K:
(a) If the original volume of the gas was 0.25 m3, what is the final volume?
(b) What is the change in volume ?V for this process?
(c) How much work does the gas do on the surroundings during the expansion?
(d) If the initial volume was 0.32 m3 and the same temperature change occurred, would the work done be the same as in the first case? Answer in Yes/No.
(e) Is the same amount of gas involved in these two situations?
Explanation / Answer
1. (a) Heat needed to convert water to boiling point that is to 100°C is given as
Q = mc?T = [0.04×4186×(100-64)] J = 6027.84 J
Where, c = specific heat of water = 4186 J/kg.°C
(b) Again heat needed to convert water at 100°C to steam at 100°C is
Q' = mL = (0.04×2256×10^3) J = 90240 J
Where L = 2256×10^3 J/kg = Latent heat of vaporization of water
Hence total heat required to convert 40 g of water at 64°C to steam at 100 °C is = Q+Q' = (6027.84+90240)
= 96267.84 J
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