course Phy 122
Ch.14 (1,2,3,7,8,9,13)1. How much heat (in joules) is required to raise the temp. of 40.0kg of water from 15ºC to 95ºC?
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95-15 = 80ºC
Q=mc∆T
30kg (4186J/kg*Cº) (80) = 10046400J
2. To what temp. will 7700 J of heat raise 3.0kg of water that’s initially at 10ºC?
Q=mc∆T 7700J=3kg(4186J/kg*Cº)∆T
7700J=12558∆T
0.613=∆T
0.613=T-10
10.613ºC
3. An average active person consumes about 2500 Cal a day. a) What’s this in joules? b) What’s this in kilo watt-hours? c) Your power company charges about a dime per kilowatt-hour. How much would your energy cost per day if you bought it from the power company? Could you feed yourself on this much $ per day?
a) 2500cal=2500kcal (4.186x10^3J/kcal) = 10465000J
b) 1kWh = 3.60x10^6J 10465000J / 3.6x10^6J = 2.9kWh
c) 0.1x2.9kWh = 29cents and no you could not feed yourself on 29 cents per day.
7. How many kilocalories are generated when the brakes are used to bring a 1200kg car to rest from a speed of 95km/h?
1km/h= 0.278m/s= 26.41m/s
1kcal=4.186x10^3J
W=1/2mv^2 =1/2 (1200kg)(26.41m/s)^2 = 418492.86
418492.86 / 4.186x10^3J = 100kcal
8. An automobile cooling system holds 16L of water. How much heat does it absorb if its’ temp. rises from 20ºC to 90ºC?
1L=1x10^3kg/m^3 16L(1x10^-3m^3)= 0.016m^3
m=pV (1x10^3kg/m^3)(0.016m^3) = 16kg
90ºC -20ºC = 70ºC Q=mc∆T
Q= 16kg(4186)(70ºC) = 4688320J
9. What’s the specific heat of a metal substance if 135kJ of heat is needed to raise 5.1kgof the metal from 18ºC to 31.5ºC?
Q=mc∆T 135kJ=135000J 31.5ºC -18ºC = 13.5ºC
135000J=5.1kg c (13.5ºC) 26470.58824J/kg = c (13.5)
= 1960.78J/kg*Cº
13. A hot iron horseshoe (mass=0.40kg) just forges is dropped into 1.35L of water in a 0.30kg iron pot initially at 20ºC. If the final equilibrium temp. is 25ºC, estimate the intial temp. of the hot horseshoe.
25 degrees C -20= 5 degrees
Q of iron pot = 0.30kg(450)(5)=657J
Q of water=1.35kg(4186)(5)=28255.5J
657+28255.5J = 28912.5J
.40kg(5ºC)28912.5J=
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