Kinetic Energy and Heat? A problem I can't figure out!?

Question

Assume that the kinetic energy of a 1400 kg car moving at 115 km/h could be converted entirely into heat. What amount of water could be heated from 20C to 50C by the car's energy? Note: 1J=1kg m^2/s^2.

Ok, I cannot figure out how to set the problem up...i dont need the answer, but just the set-up of the problem. Thanks.

*please do not refer me to chemistry.com; they are no help at all*

Answer 1

115km/hr=115*.2777m/sec
=31.9355m/sec
KE of car=1/2*m*v^2
=1/2*1400kg*31.9355^2m^2/sec^2
=713913.312kgm^2/sec^2
=713913.312J
=713.913kJ
=713.913*.2388KCal
=170.48kCal
Amount of heat reqd. to
heat 1kg of water from 20 to 50DegC
=mst
=1*1*30 kcal
=30kCal
Amount of water that can
be heated from 20 to 50DegC
=170.48/30
=56.83kg

Answer 2

I think the question should be fine, assuming that a person converts using 1/2 mv^2 to count the amount of kinetic energy released. Then from there, that amount of energy will be used to count the amount of water that can be heated in the formula that is also used for deriving the amount of heat released for the boiling and melting point of water and ice respectively. And by the way, this is more physics than chemistry in my opinion. Try checking out a physics site.

Answer 3

Start by finding the kinetic energy of the car. That's 0.5*(m_c)*v^2. m_c is the mass of the car, and v is its velocity. You'll need to convert the speed from km/h to m/s by using a conversion factor of (1000 m / 1 km)(1 hr / 3600 s).

If the kinetic energy is E, you need to solve E = (m_w)CT, where m_w is the unknown mass of water, C is the specific heat of water, and T is the change in temperature, 50 - 20 = 30 degrees Celcius. The specific heat of water is 4184 J/kg-K, where degrees Celcius and degrees Kelvin are equal as units, but you can write the units as J/kg-C if you prefer.

Answer 4

I believe K.E.= 1/2 mv^2
You have m=1400kg, v=115km/h=115000m/h=(115000/60)m/s
use the kg & m/s numbers to get the K.E. in kg m^2/s^2 which then becomes Joules.
convert Joules to calories (I'm sure you can look up the conversion)
1 calorie converts 1g of water 1degree. 30 calories converts 1g of H2O from 20 to 50 degree C. So, divide the calories by 30 and that'll give you the number of grams of water.

Answer 5

first you have to get the value of the kinetic energy of the moving car, which is given by:

KE = 1/2 (mv^2)

and then this energy is entirely converted to heat. Heat is given by

H = mc x (change in temperature)
// i guess you know the equations already

and then equate the two equations since they are of the same energy value

1/2(mv^2) = mc x (change in temperature)

then you get the mass of water!

Answer 6

Here U should now about the energy theorm.It is basic theorm for the calculations of interchaning the work and energy process.The total work done by the body is net kinetic energy.Refer www.wikipedia.org.Therefore

1/2m1v^2=m2s(t2-t1).Here m1 and m2 are different.m1 refers mass of the car.m2 refersmass of the water.Don't confuse here when there are two systems u have to identify no.of masses involved and replace in the proper formula..If u have any doubt mail me [[e-mail]]

Answer 7

1) We know that Kinetic Energy expression is:

Ek = (1/2) m1*v² .......where: m1 = mass of car

2) This amount of energy is assumed that will be converted to heat, so,

Q = (1/2) m1*v² ......(Eq.1)

3) Restriction is that temperature of water will raise from 20°C to 50°C. One expression that relates heat with temperature is the Specific Heat:

Q = m2*c*D(T) where: D(T) = temperature increase.
........................................m2 = mass of water
........................................c = specific heat of water

5) We can equate this expression with Eq.1:

m2*c*D(T) = (1/2)m1*v²

6) We can solve for m2:

m2 = (1/2)m1*v² / c*D(T)

This way you get the mass of water.

That´s it!

Good luck!