To save on heating costs, the owner of agreenhouse keeps 800.0 kg of water aroun

Question

To save on heating costs, the owner of agreenhouse keeps 800.0 kg of water around in barrels. During awinter day, the water is heated by the sun to 50.0 degreesFahrenheit. During the night the water freezes into ice at 32.0degrees Fahrenheit in 10 hours. If, instead, an electrical heatingsystem is used providing the same heating effect as thewater, a. What minimum ampere ratingwould the 240 volt circuit breaker have to be to avoidtripping? b. What would be the value of thepeak current in such a circuit? c. What would be the value of theresistance of the heating coil used? To save on heating costs, the owner of agreenhouse keeps 800.0 kg of water around in barrels. During awinter day, the water is heated by the sun to 50.0 degreesFahrenheit. During the night the water freezes into ice at 32.0degrees Fahrenheit in 10 hours. If, instead, an electrical heatingsystem is used providing the same heating effect as thewater, a. What minimum ampere ratingwould the 240 volt circuit breaker have to be to avoidtripping? b. What would be the value of thepeak current in such a circuit? c. What would be the value of theresistance of the heating coil used?

Explanation / Answer

The mass of water kept by the owner of a greenhouse is m =800.0 kg The water is heated by the sun to 50.0 degreesFahrenheit The water freezes into ice at 32.0 degrees Fahrenheit The time taken by water to freeze into ice is t = 10 hours =10 * 60 * 60 s = 36 * 103 s The heat gained by water when it is heated in the sun is Q1= m * sw* T1--------------(1) Here,sw is the specific heat of water and its valueis 4186 J/kg/K and T1= 50.0 oF = (255.93 +50.0) K = 305.93 K The heat lost by water when it freezes into ice is Q2= m * si* T2----------------(2) Here,si is the specific heat of ice and its valueis 2093 J/kg/K and T2= 32.0 oF = (255.93 +32.0) K = 287.93 K Therefore,from equations (1) and (2),the heat lost by wateris Q = Q1 - Q2 or Q = m * sw* T1 - m * si*T2 or Q = m * (sw* T1 - si*T2) a.The minimum ampere rating would the 240 volt circuit breakerhave to be to avoid tripping be I.Therefore,we get P = V * I ---------------(1) Here,P = (Q/t) or P = (m * (sw* T1 -si* T2)/t) or P = (800.0 * (4186 * 305.93 - 2093 * 287.93)/(36 *103)) or P = 15066.34 Watt From equation (1),we get I = (P/V) Here,V = 240 V or I = (15066.34/240) = 62.77 A b.Let the peak value of the current in the circuit beIo.Therefore,we get I = Io * (Exp(eV/kT) - 1) ----------------(2) Here,e = 1.6 * 10-19 C,V = 240 V,k = 1.38 *10-23 and T = (T1 - T2) = (305.93- 287.93) K = 18 K Here,(eV/kT) = (1.6 * 10-19 * 240/1.38 *10-23 * 18) = 15.46 * 104 and (Exp(eV/kT) - 1) = (Exp(15.46 * 104) - 1) =15.46e * 104 - 1 = 41.89 * 104 From equation (2),we get Io= (I/(Exp(eV/kT) - 1)) or Io= (62.77/41.89 * 104) = 1.5 *10-4 A = 0.15 * 10-3 A = 0.15 mA c.Let the value of the resistance of the heating coil used beR.Therefore,we get V = I * R or R = (V/I) or R = (240/62.77) = 3.82 Here,sw is the specific heat of water and its valueis 4186 J/kg/K and T1= 50.0 oF = (255.93 +50.0) K = 305.93 K The heat lost by water when it freezes into ice is Q2= m * si* T2----------------(2) Here,si is the specific heat of ice and its valueis 2093 J/kg/K and T2= 32.0 oF = (255.93 +32.0) K = 287.93 K Therefore,from equations (1) and (2),the heat lost by wateris Q = Q1 - Q2 or Q = m * sw* T1 - m * si*T2 or Q = m * (sw* T1 - si*T2) a.The minimum ampere rating would the 240 volt circuit breakerhave to be to avoid tripping be I.Therefore,we get P = V * I ---------------(1) Here,P = (Q/t) or P = (m * (sw* T1 -si* T2)/t) or P = (800.0 * (4186 * 305.93 - 2093 * 287.93)/(36 *103)) or P = 15066.34 Watt From equation (1),we get I = (P/V) Here,V = 240 V or I = (15066.34/240) = 62.77 A or P = (800.0 * (4186 * 305.93 - 2093 * 287.93)/(36 *103)) or P = 15066.34 Watt From equation (1),we get I = (P/V) Here,V = 240 V or I = (15066.34/240) = 62.77 A b.Let the peak value of the current in the circuit beIo.Therefore,we get I = Io * (Exp(eV/kT) - 1) ----------------(2) Here,e = 1.6 * 10-19 C,V = 240 V,k = 1.38 *10-23 and T = (T1 - T2) = (305.93- 287.93) K = 18 K Here,(eV/kT) = (1.6 * 10-19 * 240/1.38 *10-23 * 18) = 15.46 * 104 and (Exp(eV/kT) - 1) = (Exp(15.46 * 104) - 1) =15.46e * 104 - 1 = 41.89 * 104 From equation (2),we get Io= (I/(Exp(eV/kT) - 1)) or Io= (62.77/41.89 * 104) = 1.5 *10-4 A = 0.15 * 10-3 A = 0.15 mA c.Let the value of the resistance of the heating coil used beR.Therefore,we get V = I * R or R = (V/I) or R = (240/62.77) = 3.82 From equation (2),we get Io= (I/(Exp(eV/kT) - 1)) or Io= (62.77/41.89 * 104) = 1.5 *10-4 A = 0.15 * 10-3 A = 0.15 mA c.Let the value of the resistance of the heating coil used beR.Therefore,we get V = I * R or R = (V/I) or R = (240/62.77) = 3.82

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