Physics Solution Manual for 1100 and 2101

# Solution the time required to emit the energy from

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Unformatted text preview: t seconds, where Lf = 33.5 ×104 J/kg is the latent heat of fusion of water. The mass of ice per second that Chapter 13 Problems melts, then, is given by the ratio 697 m Q . The rate of heat flow through the copper rod is t t Q kA ∆T = (Equation 13.1), where k is the thermal conductivity of copper, A t L and L are, respectively, the cross-sectional area and length of the rod, and ∆T = 100.0 C° is the difference in temperature between the boiling water and the ice-water mixture. found from Q . Dividing this by the Lf elapsed time t, we obtain an expression for the mass of ice per second that melts: SOLUTION Solving Q = mLf (Equation 12.5) for m yields m = Q m t = t Lf Substituting (1) Q kA ∆T = (Equation 13.1) into Equation (1), we find that t L m kA ∆T 390 J/ ( s ⋅ m ⋅ Co ) ( 4.0 ×10−4 m2 ) (100.0 Co ) = = = 3.1×10−5 kg/s 4 t Lf L (1.5 m ) ( 33.5 ×10 J/kg ) 15. REASONING Heat is delivered from the heating element to the water via conduction. The amount of heat Q conducted in a time t is given by Q= ( kcopper A∆T ) t (13.1) L where kcopper is the thermal conductivity of copper, A is the area of the bottom of the pot, ∆T is the temperature difference, and L is the thickness of the bottom of the pot. Since the water is boiling under one atmosphere of pressure, the temperature difference is ∆T = TE – 100.0 °C, where TE is the temperature of the heating element. Substituting this expression for ∆T into Equation (13.1) and solving for TE, we have TE = 100.0 °C + LQ kCopper At (1) When water boils, it changes from the liquid to the vapor phase. The heat required to make the water change phase is Q = mLv, according to Equation 12.5, where m is the mass and Lv is the latent heat of vaporization of water. 698 THE TRANSFER OF HEAT SOLUTION Substituting Q = mLv into Equation (1), and noting that the bottom of the pot is circular so that its area is A = π R , we have that 2 TE = 100.0 °C + LQ kCopper At = 100.0 °C + L ( mLv ) () kCopper π r 2 t (2) The thermal conductivity of copper can be found in Table 13.1 kcopper = 390 J/ ( s ⋅ m ⋅ C° ) , and the latent heat of vaporization for water can be found in Table 12.3 5 (Lv = 22.6 × 10 J/kg). The temperature of the heating element is TE ( 2.0 × 10−3 m ) ( 0.45 kg ) ( 22.6 × 105 J/kg ) = = 100.0 °C + 390 J/ ( s ⋅ m ⋅ C°) π ( 0.065 m ) (120 s ) 2 103.3 °C 16. REASONING Heat Q flows along the length L of the bar via conduction, so that ( k A∆T ) t , where k is the thermal conductivity of the material Equation 13.1 applies: Q = L from which the bar is made, A is the cross-sectional area of the bar, ∆T is the difference in temperature between the ends of the bar, and t is the time during which the heat flows. We will apply this expression twice in determining the length of the bar. SOLUTION Solving Equation 13.1 for the length L of the bar gives L= ( k A∆T ) t = k A ( TW − TC ) t Q Q (1) where TW and TC, respectively are the temperatures at the warmer and cooler ends of the bar. In this result, we do not know the terms k, A, t, or Q. However, we can evaluate the heat Q by recognizing that it flows through the entire length of the bar. This means that we can also apply Equation 13.1 to the 0.13 m of the bar at its cooler end and thereby obtain an expression for Q: k A ( T − TC ) t Q= D where the length of the bar through which the heat flows is D = 0.13 m and the temperature at the 0.13-m point is T = 23 °C, so that ∆T = T − TC . Substituting this result into Equation (1) and noting that the terms k, A, and t can be eliminated algebraically, we find Chapter 13 Problems L= k A ( TW − TC ) t Q = k A ( TW − TC ) t k A ( T − TC ) t = 699 k A ( TW − TC ) t D k A ( T − TC ) t D = ( TW − TC ) D = ( 48 °C − 11 °C )( 0.13 m ) = 0.40 m 23 °C − 11 °C ( T − TC ) 17. REASONING The heat Q required to change liquid water at 100.0 °C into steam at 100.0 °C is given by the relation Q = mLv (Equation 12.5), where m is the mass of the water and Lv is the latent heat of vaporization. The heat required to vaporize the water is conducted through the bottom of the pot and the stainless steel plate. The amount of heat ( k A ∆T ) t (Equation 13.1), where k is the thermal conducted in a time t is given by Q = L conductivity, A and L are the cross-sectional area and length, and ∆T is the temperature difference. We will use these two relations to find the temperatures at the aluminum-steel interface and at the steel surface in contact with the heating element. SOLUTION a. Substituting Equation 12.5 into Equation 13.1 and solving for ∆T, we have ∆T = QL ( mLv ) L = k At k At The thermal conductivity kAl of aluminum can be found in Table 13.1, and the latent heat of vaporization for water can be found in Table 12.3. The temperature difference ∆TAl between the aluminum surfaces is ∆TAl ( mLv ) L = ( 0.15 kg ) ( 22.6 ×105 J/kg ) ( 3.1×10−3 m ) = 1.2 C° = kAl At ( ) 240 J/ ( s ⋅ m ⋅ C° ) 0.015 m 2 ( 240 s ) The temperature at the aluminum-steel interface is TAl-Steel = 100.0 °C + ∆TAl = 101.2 °C . b. Using the thermal condu...
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