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Unformatted text preview: cs m∆Ts = m ( cw ∆Tw + Lf + cs ∆Ts ) (2) The snow maker pumps and sprays lake water at a rate of 130 kg per minute, so the mass m
of the water turned to snow in one minute is m = 130 kg. Equation (2) then yields the total
amount of heat transferred to the air each minute: 666 TEMPERATURE AND HEAT { ( )( ) ( )( Q = (130 kg ) 4186 J/ kg ⋅ Co 12.0 C o + 33.5 × 104 J/kg + 2.00 × 103 J/ kg ⋅ Co 7.0 C o )} = 5.2 × 107 J 67. SSM REASONING According to the statement of the problem, the initial state of the
system is comprised of the ice and the steam. From the principle of energy conservation,
the heat lost by the steam equals the heat gained by the ice, or Qsteam = Qice . When the ice
and the steam are brought together, the steam immediately begins losing heat to the ice. An
amount Q1(lost) is released as the temperature of the steam drops from 130 °C to 100 °C, the
boiling point of water. Then an amount of heat Q2(lost) is released as the steam condenses
into liquid water at 100 °C. The remainder of the heat lost by the "steam" Q3(lost) is the heat
that is released as the water at 100 °C cools to the equilibrium temperature of Teq = 50.0 °C .
According to Equation 12.4, Q1(lost) and Q3(lost) are given by
Q1(lost) = csteam msteam (Tsteam − 100.0 °C) and Q3(lost) = cwater msteam (100.0 °C – Teq )
Q2(lost) is given by Q2(lost) = msteam Lv , where Lv is the latent heat of vaporization of water.
The total heat lost by the steam has three effects on the ice. First, a portion of this heat
Q1(gained) is used to raise the temperature of the ice to its melting point at 0.00 °C. Then, an amount of heat Q2(gained) is used to melt the ice completely (we know this because the
problem states that after thermal equilibrium is reached the liquid phase is present at
50.0 °C). The remainder of the heat Q3(gained) gained by the "ice" is used to raise the
temperature of the resulting liquid at 0.0 °C to the final equilibrium temperature. According
to Equation 12.4, Q1(gained) and Q3(gained) are given by
Q1(gained) = cice mice (0.00 °C – Tice ) and Q3(gained) = cwater mice (Teq – 0.00 °C)
Q2(gained) is given by Q2(gained) = mice Lf , where Lf is the latent heat of fusion of ice. SOLUTION According to the principle of energy conservation, we have
Qsteam = Qice
Q1(lost) + Q2(lost) + Q3(lost) = Q1(gained) + Q2(gained) + Q3(gained) Chapter 12 Problems 667 or
csteam msteam (Tsteam − 100.0 °C) + msteam Lv + cwater msteam (100.0 °C – Teq )
= cice mice (0.00 °C –Tice ) + mice Lf + cwater mice (Teq – 0.00 °C) Values for specific heats are given in Table 12.2, and values for the latent heats are given in
Table 12.3. Solving for the ratio of the masses gives
msteam
mice = cice (0.00 °C –Tice ) + Lf + cwater (Teq – 0.00 °C)
csteam (Tsteam − 100.0 °C) + Lv + cwater (100.0 °C – Teq ) 3
4 2.00 ×10 J/ ( kg ⋅ C° ) [ 0.0 °C–(–10.0°C) ] + 33.5 ×10 J/kg + 4186 J/ ( kg ⋅ C° ) ) (50.0 °C–0.0 °C)
=
5 2020 J/ ( kg ⋅ C° ) (130 °C − 100.0 °C) + 22.6 ×10 J/kg + 4186 J/ ( kg ⋅ C° ) (100.0 °C–50.0 °C) or msteam = 0.223
mice
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68. REASONING When the minimum mass mw of water is used to solidify the molten gold,
the heat Qg lost by the gold is exactly the amount needed to raise the temperature of the
water to the boiling point and then convert the water into steam. If the mass of the water is
smaller than mw, all of the water will be converted to steam before the gold has entirely
solidified. The water first gains an amount of heat Q1 = cw mw ∆T (Equation 12.4) when it is raised to the boiling point (100.0 °C), where cw = 4186 J/ ( kg ⋅ Co ) is the specific heat
capacity of water (see Table 12.2) and ∆T is the change in its temperature. When the boiling
water is converted to steam, it absorbs a further amount of heat Q2 = mw Lw (Equation 12.5),
where Lw = 22.6×105 J/kg is the latent heat of vaporization of water (see Table 12.3). The
total amount of heat given off by the molten gold as it solidifies is found from Qg = mg Lg
(Equation 12.5), where mg is the mass of the gold and Lg = 6.28×104 J/kg is the latent heat
of fusion of gold (see Table 12.3). The total amount of heat Qw gained by the water is equal
to the heat lost by the gold, so we have that
Qw = Q1 + Q2 = Qg (1) 668 TEMPERATURE AND HEAT SOLUTION Substituting Q1 = cw mw ∆T (Equation 12.4), Q2 = mw Lw (Equation 12.5), and
Qg = mg Lg (Equation 12.5) into Equation (1), we obtain
cw mw ∆T + mw Lw = mg Lg
144 2444
4
3
{ Heat gained by water (2) Heat lost
by gold Solving Equation (2) for mw yields
mw ( cw ∆T + Lw ) = mg Lg mw = or mg Lg
cw ∆T + Lw Therefore, mw = 69. ( 0.180 kg ) ( 6.28 ×104 J/kg )
= 4186 J/ ( kg ⋅ Co ) (100.0 oC − 23.0 oC ) + 22.6 ×105 J/kg 4.38 ×10−3 kg SSM WWW REASONING The system is comprised of the unknown material, the
glycerin, and the aluminum calorimeter. From the principle of energy conservation, the heat
gain...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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