anonymous
  • anonymous
The specific heat capacity of water= 4200 J kg-1K-1 1) How much heat energy is required to warm a) 1 kg of water from 14°C to 15°C; b) 5 kg of water from 14°C to 15°C; and c) 5 kg of water from 14°C to 22°C. THANK YOU!!
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[J kg ^{-1} K ^{-1}\] this is the correct way of writing the unit
AravindG
  • AravindG
Use the formul heat required =\(mc \Delta t\)
chmvijay
  • chmvijay
q=McDelta T

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

AravindG
  • AravindG
where m is mass c is specific heat capacity ,delta t=change in temperature
anonymous
  • anonymous
ok thanks
AravindG
  • AravindG
yw :)
anonymous
  • anonymous
i worked it out...is the answer 4200J
AravindG
  • AravindG
a b or c?
anonymous
  • anonymous
for a)
AravindG
  • AravindG
right !
anonymous
  • anonymous
hmm and b) is 21000J while c) is 1.68*10\[1.68\times10^{5}\] am i right???
AravindG
  • AravindG
yep :)
anonymous
  • anonymous
was it that simple?
anonymous
  • anonymous
question 2! one way to measure the energy contained in a beam of x-rays is to absorb the beam by a piece of pure graphite. If x-rays are shone on to a cavity in a block of graphite of mass 10 g (0.01 kg), and a rise in temperature of 0.005K, work out the energy supplied by the beam.
anonymous
  • anonymous
oh and the specific heat capacity of graphite is 600\[J kg ^{-1}K ^{-1}\]
AravindG
  • AravindG
same formula !
anonymous
  • anonymous
ok......just making sure. question 3! An electric drill consuming energy at a rate of 200 W is used to make a small hole in a block of copper of mass 2 kg. Assuming that 80% of the electrical energy used ends up in the block as heat, and that this heat is spread evenly throughout the block, estimate its rise in temperature in 10 seconds.
austinL
  • austinL
Why dont you make a new question? rather than lump several into one post?
anonymous
  • anonymous
ok @kym02 i'll give a brief info. so energy is 200 W i.e. it uses 200 J in 1 second, correct?
anonymous
  • anonymous
um i honestly dont know....im a beginner. Dont really know about this stuff! sorry!
anonymous
  • anonymous
Ok, then there is one concept. W means Watt. Watt is the unit of power of a machine. We say that the power of a machine is the amount of work it can do in 1 second.(i.e. rate of doing work,) Hence its unit is Watt(W) also expressed as Joule/second. so 200 W basically means that the machine will use 200 Joules of energy in 1 second. Ok?
anonymous
  • anonymous
ok i understand.
anonymous
  • anonymous
Fine. so if the machine uses 200 Joules of energy in 1 second, how much of energy would it use in 10 seconds?
anonymous
  • anonymous
2000 J.
anonymous
  • anonymous
Ok. 2000 J is the electrical energy used. Now he says that 80% of this 2000 J is used as heat energy. So what is 80% of 2000 ?
anonymous
  • anonymous
1600
anonymous
  • anonymous
Thats right. So this 1600 is the heat energy. As shown above, Heat energy = Mass x sp. heat capacity x rise in temp. You have mass = 2 kg, sp. heat capacity(c) = 400 JKg^-1K^-1, can you find the rise in temperature?
anonymous
  • anonymous
i dont know that one, do i have to transpose?
anonymous
  • anonymous
Yep you have to. above there was a formula mentioned Heat energy = m x c x delta T Here Heat energy = 1600, m = 2 kg, c = 400 J/KgK, u need to find delta T
anonymous
  • anonymous
m means Mass; c means specific heat capacity and delta T means the rise in temperature.
anonymous
  • anonymous
\[E _{H}=mcΔT\] so the temperature rise will be: EH divided by mass times c am i right?
anonymous
  • anonymous
Thats perfectly correct. Just type d ans also when you find it. :)
anonymous
  • anonymous
ok i will, im working on it
anonymous
  • anonymous
Is the answer 2°C?
anonymous
  • anonymous
Yes. :)
anonymous
  • anonymous
yes!!!!! thanks :) i have one more question though
anonymous
  • anonymous
a)How much potential energy does a 2.0 kg lump of copper possess when it is lifted 3.0 m from the ground? (Take g to be 10 N kg^-1.) b) If it is dropped, how much kinetic energy will it have just before it hits the ground? c) If all that energy is converted to heat, and if all that heat goes into the copper, what rise in temperature would you expect? That's all ! :)
anonymous
  • anonymous
a) m = 2, h = 3, g = 10. Use Potential Energy = mgh. b) Potential Energy found in a) = Kinetic Energy (By law of conservation of energy) c) Whatever PE you get, assume that as heat energy, the use C , M of copper to find rise in temp by formula EH = m.c. delta T.
anonymous
  • anonymous
whats the unit for a) ???
anonymous
  • anonymous
a) Any form of energy, whether potential energy, electric energy, heat energy, is expressed in joules( J ) (Please please note this fact; )
anonymous
  • anonymous
ok ok so a) =60 J
anonymous
  • anonymous
Thats correct.
anonymous
  • anonymous
and c) = \[7.5\times 10^{-2} \deg C\]
anonymous
  • anonymous
Which is correct, :)
anonymous
  • anonymous
I hope you're familiar with law of conservation of energy right?
anonymous
  • anonymous
no but i will just look back in a textbook! thank you soooo much!!! you saved me!! :)
anonymous
  • anonymous
yeah no probs as long as you're familiar with concepts and numericals :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.