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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
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\[J kg ^{-1} K ^{-1}\] this is the correct way of writing the unit
Use the formul heat required =\(mc \Delta t\)
q=McDelta T

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where m is mass c is specific heat capacity ,delta t=change in temperature
ok thanks
yw :)
i worked it out...is the answer 4200J
a b or c?
for a)
right !
hmm and b) is 21000J while c) is 1.68*10\[1.68\times10^{5}\] am i right???
yep :)
was it that simple?
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.
oh and the specific heat capacity of graphite is 600\[J kg ^{-1}K ^{-1}\]
same formula !
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.
Why dont you make a new question? rather than lump several into one post?
ok @kym02 i'll give a brief info. so energy is 200 W i.e. it uses 200 J in 1 second, correct?
um i honestly dont know....im a beginner. Dont really know about this stuff! sorry!
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?
ok i understand.
Fine. so if the machine uses 200 Joules of energy in 1 second, how much of energy would it use in 10 seconds?
2000 J.
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 ?
1600
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?
i dont know that one, do i have to transpose?
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
m means Mass; c means specific heat capacity and delta T means the rise in temperature.
\[E _{H}=mcΔT\] so the temperature rise will be: EH divided by mass times c am i right?
Thats perfectly correct. Just type d ans also when you find it. :)
ok i will, im working on it
Is the answer 2°C?
Yes. :)
yes!!!!! thanks :) i have one more question though
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 ! :)
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.
whats the unit for a) ???
a) Any form of energy, whether potential energy, electric energy, heat energy, is expressed in joules( J ) (Please please note this fact; )
ok ok so a) =60 J
Thats correct.
and c) = \[7.5\times 10^{-2} \deg C\]
Which is correct, :)
I hope you're familiar with law of conservation of energy right?
no but i will just look back in a textbook! thank you soooo much!!! you saved me!! :)
yeah no probs as long as you're familiar with concepts and numericals :)

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