perature of 200 g of w ater from 10°C to the boiling point and to vaporize 10% of it? Assume th a t there are no energy los ses.
Given: m w = 0.2 kg is the m ass of the cold w ater, t = 10°C, or T = 283 K , is the tem perature of the cold w ater, ttisteam = 0.1 m w is the m ass of steam . From tab les, we find the specific h eat of w ater, cw = 4187 J /(k g -K ) ~ 4190 J /(k g -K ), the boiling po in t of w ater, t b = 100°C, or T h = 373 K , and the specific la te n t heat of vaporization of w ater, r = 2.26 x 106 J/k g .
F in d : th e am ount of h e a t Q (spent energy).
S olu tio n . Since the w ater tem perature is below the boiling point, it m u st be heated from T to T b, for w hich the am ount Qx = cwm w ( T h — T) of h e a t is required. The am ount of h eat required to vaporize the w ater is Q2 = w steamr * or Q2 = 0.1 /rcwr.
The process is known to occur w ith o u t energy losses. Consequently, the to ta l am ount of heat th a t m ust be spent is
Q = Qi + (?2>
Qx = 4190 J/(k g -K ) x 0.2 kg x 90 K = 75420 J = 75.42 k J,
§ 5. Properties of Vapours 53 Q2 = 0.1
x
0.2 kgx
2.26x
106 J/k g = 4.52x
104 J= 45.2 k J,
Q = 75.42 kJ + 45.2 kJ = 120.62 k J.
This process can be represented graphically (Fig. 9). We shall p lo t the am ount of spent heat along the abscissa axis. For the sake of sim plicity, we
round off the values of Qx and Q2 to integers: 75 and 45 k J. The 700' tem perature in degrees Celsius (this is more convenient for cho osing a scale) is plotted along the o rdinate axis. W hen analyzing the graph, we pay atte n tio n to the fact th a t vaporization occurs 10
a t a co n stant tem perature (in th is o I T k T case, a t 100°C). Consequently, the
water m ust be heated to th is tern- Fig- 9 perature. The intercepts OA and
A B on th e abscissa correspond to the values Qx and Q2. Answer. A bout 121 kJ of energy is required to heat the w ater and vaporize p art of it.
Problem 27. How much charcoal m ust be b u rn t in order to heat 6 t of w ater taken a t 283 K to boiling and vaporize 1 t of it? The efficiency of the boiler is 70% .
Given: raw = 6 X 103 kg is the mass of w ater in the boiler, T = 283 K is the in itia l tem perature of the w ater, T h = 373 K is the boiling point of w ater, m sieam = 103 kg is the mass of the steam , and r| = 70% = 0.7 is the efficiency of the boiler. From tables, we take the specific heat of w a ter, c = 4187 J/(k g -K ) ~ 4190 J/(k g -K ), the specific la te n t h eat of vaporization of w ater, r = 2.26 X 106 J/k g , and specific h eat of com bustion of charcoal, q — 3.1 X 107 J /k g .
F in d : the mass m c of charcoal.
Solu tio n . The energy Q = qmc liberated during th e com bustion of the charcoal is spent on heating the w ater to b o il ing point, Qx = cmw ( T h — T), and on vaporizing 1 t of the w ater, Q2 = rm steam-
From the energy conservation law and given th a t only 70% of energy lib erated by burning the charcoal is spent on heating and vaporizing the w ater, we w rite the heat balance equation v\Q = Qx + Q2, or v\qmQ = cmw ( T h — T) +
54 Ch. I. Fundam entals of M olecular P h ysics Whence ( r b - r ) + m steam _
S u b stitu tin g the num erical values, we obtain
4190 J / (kg-K ) X 6 X 103 k g x 9 0 K + 2.26 x 106 J/k g X 103 kg 0.7 X 3.1 X 1 0 7 J/k g “ = 208.
Answer. A bout 208 kg of charcoal are spent.
Problem 28* 200 kg of steam a t a tem perature of 373 K are
passed through 4 t of w ater a t a tem perature of 293 K . To w h at tem peratu re w ill the w ater be heated? Energy losses should be neglected. G raph th e function t = / (Q).
Given: raw = 4
X
103 kg is the mass of the w ater, T w = 293 K is th e tem perature of the w ater, msteam = 200 kg is the m ass of the steam , r 6team = 373 K is the tem perature of the steam . From tables, we take the specific h eat of w ater, cw = 4187 J/(k g -K ) ~ 4190 J/(kg*K ), and th e specific la te n t h e a t of vaporization of w ater, r = 2.26X
106 J/k g .Find: th e tem perature 0 a t the end of the process. Solution. W e have here a h eat exchange: steam a t its boil ing p o in t (which equals its condensation tem perature) gives to w ater the am ount of heat Qt = rrasteam and is converted into w ater a t the same tem p eratu re (during condensation, as in boiling, the tem perature rem ains constant). The w ater obtained from the steam is cooled from T h to 0 , lib e ratin g th e am ount of heat Q2 = Cw^steam (^steam — ©)• Since the final tem perature 0 is the same for all com ponents, th e in tern al energy of the cold w ater increases by Q = cwm w X
(e - rw).
According to the energy conservation law, we can w rite
@1 4 “
Q
2 =Qi
o r f^ s t e a m 4 “ ^w ^steam (^ s te a m= cwm w (0 — T w). We transform th e heat balance equation rem oving the pa rentheses and gathering the term s containing the unknown tem perature on the right-hand side:
r tf ls te a m C w ^ s te a m ^ 6 te a m 4 " Cw fflw Tw — CwW w0
4~ c w ^ s te a m © = ( ^ w ^ w 4 " ^ w ^ s t e a r n )
whence
q __ r/yi8team ~f~ cw (msteam^steam~Hmw^tw )
§ 5. Properties of Vapours 55 S u b stitu tin g the num erical values, we determ ine 0 :
0 = 2 .2 6 x 106 J/kg x 200 kg 4190 J/(kg*K ) (4 x 103 k g + 2 0 0 kg)
4190 J /(k g -K ) (200 kg
x
373 K + 4x
103 kgx
293 K) _ o o o i f + 4190 J/(k g-K ) (4x
103 k g + 200 kg) ~ ^ A or 0 = 49°C.Figure 10 shows the tem perature versus the am ount of heat, t = f (iQ). The tem perature rem ains unchanged over segment A B (the evolution of h eat during condensation occurs due to a decrease in the poten
tia l energy of interaction between the m olecules). The tem perature of condensate falls from ^steam to © 1 over segm ent BC (cooling the con densate lib erates heat). The tem per- A atu re of th e cold w ater increases from T to 0 over segment CD (the < process involves the absorption of heat).
Answer. The final tem perature 0 Fi£- 10 is ap p rox im ately 49°C.