Calories generated in brakes of a machine

1. Oct 11, 2014

Karol

1. The problem statement, all variables and given/known data
A 1000 kg mass car is driving at 3 [m/sec], how many kcal develop in the brakes till full stop

2. Relevant equations
Kinetic energy: $E=\frac{1}{2}mv^2$
Energetic equivalent of a kcal: 4186 [joule]

3. The attempt at a solution
$E=\frac{1}{2}1000\cdot 9=4500[joule]$
$\frac{4500}{4186}=1.07[kcal]$
It should be 1000 times bigger, 1,075[kcal]

2. Oct 11, 2014

haruspex

I get your answer. Is there perhaps a confusion between decimal comma and decimal point?

3. Oct 11, 2014

Karol

by you get my answer you mean you also get the same result? i don't know if there's a confusion between a comma and a decimal point, but please tell me if the result of 1 kcal is reasonable? it isn't, right? so my answer isn't correct

4. Oct 11, 2014

haruspex

I mean I agree with your answer, a bit over one kcal. But in continental Europe, that could be written 1,075, no?

5. Oct 12, 2014

Karol

I don't know about europe, i'm interested in my result. 1.07 kcal seems to me too low for that mass and speed, isn't it? and 1075 cal are gram calories, the heat needed to raise the temperature of one gram of water. my result is indeed 1075 gram calories, but i ask about kg calories, kcal. i think the answer must be 1075 kcal, it seems more reasonable

6. Oct 12, 2014

haruspex

Bear in mind that I do not know where you are or where your textbooks come from. Many on these forums are in continental Europe. I'm trying to understand why you are told the right answer is "1,075 kcal", when you and I both calculate it as just over 1 kcal.
No, 1 kcal seems reasonable to me. 3m/s is not very fast. I would not expect the generated heat to be enough to boil off a litre of water![/QUOTE]

7. Oct 12, 2014

Karol

You are probably right. but to boil 1 liter, from 00 to 1000 we need only 100 kcal, so 1075 is much more than to boil 1 liter of water, no?