Calculating Heat Energy and Speed: Physics Final Review

[physicszman]

If you guys can check the math on these last 3 questions I would be very greatful. Sorry to throw this many questions at you but I've been doing physics 3 days straight now reviewing for my final tomorrow. Any help would be awsome.

1. How much heat energy is required to convert a 1.0kg block of ice at 0C to steam at 100C. The specific heat of water is 4.19x10^3 J/Kg. The heat of vaporization of water is 2256 x 10^3 K/kg-C. The heat of fusion of water is 333x10^3 J/kg.

Q = mc deltaT
(mF) + (mc) x (100) + (mV)
(1.0kg)(333x10^3 J/kg) + (1.0kg)(4.19x10^3 J/kg-C)(100C) + (1.0kg)(2256x10^3 J/kg)
= 9.45x10^11J
=9.45x10^11J / 4.2
=2.25^11 cal?

2. FInd the speed at which the kinteic energy of a 1000kg automobile would equal the thermal energy required to heat a 1.0kg block of ice at 0C to 100C.

mc deltaT= 1/2kv^2
(1.0kg)(4.19x10^3 J/kg-C)(100)=(500kg)v^2
=sqrt((1.0kg)(4.19x10^3 J/kg-C)(100)/(500kg))=v
= 28.94 (sorry i don't know what unit)

3. The 200 ft^2 ceiling of a room is insulated from the attick by a 6.0 inch layer of glass wool for which k = 0.04 W/m-C. If the attick temperature is 50C, adn the room temperature is maintained at 20C, how much energy per hour is transported through the insulation by conduction? Assume the drop in temperature is maintained across the glass wool.

3) heat generated=k*(Area/thickness)*(delta T)
= (.04 W/m-C)(60.96m^2/.15m)(30C)
=487.68J x 3600
=1.75x10^6J/hour


Thanks

 

[Inspector Gadget]

#1, your method is correct, but you made an error doing the math, as I got 3008000J.

#2, if you start with a kg of ice, I think it first needs to turn to water, then raise 100 degrees. Also, your final units for the velocity will be m/s.

 

[M98Ranger]

1. Your calculation for the heat energy required is correct. However, when converting to calories, it should be 2.25x10^11 cal instead of 2.25^11 cal. Also, it is more accurate to use the specific heat of water at constant pressure (4.18 J/g-C) instead of at constant volume (4.19 J/g-C). So the final answer should be 2.26x10^11 cal.

2. Your calculation for the speed is correct. The unit for speed would be m/s.

3. Your calculation for the heat transfer is correct. However, the unit for heat transfer is watts (W) or joules per second (J/s), not joules per hour (J/h). So the final answer should be 1.75x10^6 J/s or 1.75 MW.