Question

Can you help me with this question? (I will add it as a picture) (I don't need the first question but please help me with the others)

Answer 1

what do you know about the internal energy?

Answer 2

It is the sum of the total kinetic and total potential energy of a substance.

Answer 3

better is the sum of potential and kinetic energy of the particles which compose that substance

Answer 4

okay then

Answer 5

it is the definition of internal energy

Answer 6

so we answered to the question a)

Answer 7

what should I say for b

Answer 8

when we have a melting solid, we provide to such solid a flow of heat, nevertheless its temperature doesn't change

Answer 9

so, the measure of the temperature, can not be a measure of the energy received by a melting solid

Answer 10

oh I see, we add as much heat as the product of its mass and latent heat of fusion so the temperature never changes

Answer 11

that's right!

Answer 12

only after the solid is completely melted, the temperature will start to increase

Answer 13

and what about the next one

Answer 14

I'm thinking...

Answer 15

is the container closed

Answer 16

I think so! Am I right?

Answer 17

I don't know it doesn't give that much detail in the question but I think so too

Answer 18

But it also says that the container is uninsulated

Answer 19

so it does have a contact with its surroundings, therefore, energy loss

Answer 20

I think that the pressure exerted by the vapor which comes from heating, prevent the boiling of the liquid

Answer 21

prevents*

Answer 22

I don't know but it might also have something to do with the energy loss

Answer 23

It wouldn't say it is uninsulated if that information didn't really matter

Answer 24

I think that we have no energy losses

Answer 25

but the container is also in touch with the open air so it gets colder by time

Answer 26

I think that we have to explain why the liquid never reaches its boiling point

Answer 27

and I say that because of the energy loss, it never gets the required energy to get to its boiling point

Answer 28

yours sound more detailed and since we are not that high level, this might need a simpler answer, but still I don't say that I am sure

Answer 29

I think that inside the container there is a great pressure

Answer 30

okay then

Answer 31

part of the provided amount of heat goes to the vapor being produced by the the heating
and such heat increases the internal energy of such vapor

Answer 32

okay, and the last question

Answer 33

please wait, I'm thinking...

Answer 34

we have the subsequent equations:
\[\Large Q = C \cdot \Delta T\]
is the amount of heat added to the liquid
then:
\[\Large \frac{{\Delta T}}{{\Delta t}} = - 3.1\]
which comes from the text of the problem

Answer 35

isn't it like \[Q=mc DeltaT\]

Answer 36

yes! And the product \(mc\) is the heat capacity \(C\)
next we can write:

\[\Large \frac{Q}{{\Delta t}} = C \cdot \frac{{\Delta T}}{{\Delta t}} \Rightarrow C = \frac{{Q/\Delta t}}{{\Delta T/\Delta t}}\]
therefore:
\[\Large c = \frac{1}{m}\left( {\frac{{Q/\Delta t}}{{\Delta T/\Delta t}}} \right)\]

Answer 37

now the quantity:
\[\Large {Q/\Delta t}\]
is an energy over time, namely it is equal to the power added to the liquid, so we have:
\[\large c = \frac{1}{m}\left( {\frac{{Q/\Delta t}}{{\Delta T/\Delta t}}} \right) = \frac{1}{{0.240}}\left( {\frac{{35}}{{3.1/60}}} \right) = ...{\text{watts/}}\left( {{\text{Kg}} \cdot {\text{K}} \cdot {\text{se}}{{\text{c}}^{ - {\text{1}}}}} \right)\]

Answer 38

\[\large \begin{gathered}
c = \frac{1}{m}\left( {\frac{{Q/\Delta t}}{{\Delta T/\Delta t}}} \right) = \hfill \\
\hfill \\
= \frac{1}{{0.240}}\left( {\frac{{35}}{{3.1/60}}} \right) = ...{\text{watts/}}\left( {{\text{Kg}} \cdot {\text{K}} \cdot {\text{se}}{{\text{c}}^{ - {\text{1}}}}} \right) \hfill \\
\end{gathered} \]

Answer 39

but since it's asking for its specific heat capacity, the unit of the answer should be in Jkg^-1K^-1

Answer 40

yes! Please note that:
\(joules= watts \times seconds\)

Answer 41

we can write this:
\[\Large \begin{gathered}
{\text{watts/}}\left( {{\text{Kg}} \cdot {\text{K}} \cdot {\text{se}}{{\text{c}}^{ - {\text{1}}}}} \right) = \frac{{{\text{watts}}}}{{{\text{Kg}} \cdot {\text{K}} \cdot {\text{se}}{{\text{c}}^{ - {\text{1}}}}}} = \hfill \\
\hfill \\
= \frac{{{\text{watts}} \cdot {\text{sec}}}}{{{\text{Kg}} \cdot {\text{K}}}} = \frac{{{\text{joules}}}}{{{\text{Kg}} \cdot {\text{K}}}} \hfill \\
\end{gathered} \]

Answer 42

oh thanks then, it equals to 2822.58 Jkg^-1K^-1

Answer 43

that's right!

Answer 44

thanks a lot

Answer 45

:)