- #1
dustybray
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I'm not sure if I'm approaching these problems correctly. Could someone check my work?
Steam at 100°C is condensed in 500g of water at 20°C.
a. What is the maximum amount of steam that can be condensed in this amount of water?
Q[steam] = m[steam] * L[c steam] * (T[f] – T)
Q[steam] = m * 2*10^4 * (20 - 100)
Q[steam] = m * -1.6*10^6 J
Q[water] = m[water] * L[v water] * (T[f] – T)
Q[water] = .5kg * 3.33*10^5 * (100°C – 20°C)
Q[water] = 1.32*10^7
Q[steam] + Q[water] = 0
m * -1.6*10^6 J + 1.32*10^7 = 0
m = (- 1.32*10^7) / (-1.6*10^6 J)
m = 8.25kg
I don't even know what a reasonable answer would be...
b. If only 5g of steam are condensed, what is the final temperature of the water?
Q[steam] = m[steam] * c[steam] * (T[f] – T)
Q[steam] = .005kg * 2*10^4 * (T – 100)
Q[steam] = 100*T – 1*10^4
Q[water] = m[water] * c[water] * (T[f] – T)
Q[water] = .5kg * 3.33*10^5 * (T – 20°C)
Q[water] = 1.67*10^5 * T – 3.33*10^6
Q[steam] + Q[water] = 0
100*T – 1*10^4 + 1.67*10^5 * T – 3.33*10^6 = 0
100 * T + 1.67*10^5 * T = 1*10^4 + 3.33*10^6
T (100 + 1.67*10^5) = 1*10^4 + 3.33*10^6
T = (1*10^4 + 3.33*10^6) / (100 + 1.67*10^5)
T = 19.99°C
(I know most of the units are missing-- I'm just lazy... :)
Thanks,
dusty...
Steam at 100°C is condensed in 500g of water at 20°C.
a. What is the maximum amount of steam that can be condensed in this amount of water?
Q[steam] = m[steam] * L[c steam] * (T[f] – T)
Q[steam] = m * 2*10^4 * (20 - 100)
Q[steam] = m * -1.6*10^6 J
Q[water] = m[water] * L[v water] * (T[f] – T)
Q[water] = .5kg * 3.33*10^5 * (100°C – 20°C)
Q[water] = 1.32*10^7
Q[steam] + Q[water] = 0
m * -1.6*10^6 J + 1.32*10^7 = 0
m = (- 1.32*10^7) / (-1.6*10^6 J)
m = 8.25kg
I don't even know what a reasonable answer would be...
b. If only 5g of steam are condensed, what is the final temperature of the water?
Q[steam] = m[steam] * c[steam] * (T[f] – T)
Q[steam] = .005kg * 2*10^4 * (T – 100)
Q[steam] = 100*T – 1*10^4
Q[water] = m[water] * c[water] * (T[f] – T)
Q[water] = .5kg * 3.33*10^5 * (T – 20°C)
Q[water] = 1.67*10^5 * T – 3.33*10^6
Q[steam] + Q[water] = 0
100*T – 1*10^4 + 1.67*10^5 * T – 3.33*10^6 = 0
100 * T + 1.67*10^5 * T = 1*10^4 + 3.33*10^6
T (100 + 1.67*10^5) = 1*10^4 + 3.33*10^6
T = (1*10^4 + 3.33*10^6) / (100 + 1.67*10^5)
T = 19.99°C
(I know most of the units are missing-- I'm just lazy... :)
Thanks,
dusty...