Problem involving specific heat and graph

  • Thread starter frkCarl
  • Start date
  • #1
4
0

Homework Statement


A researcher studies the heat transference between an iron block and an unknown substance, inside an adiabatic calorimeter. In successive experiences he puts inside the calorimeter the unknown substance, always in its solid state and with temperature θs = 20ºC, and the iron block, in each case with a different initial temperature "θi". In each case he measures the equilibrium temperature "θe". The results are shown in the graphic.

The quotient of the mass of the iron block and the mass of the unknown substance, "mi/ms", is equal to 0,8.
Consider the specific heat of the iron = 0.1 cal/g ºC

Image:
https://imgur.com/a/QLu3NQI

Question: Find the specific heat, "cs", of the unknown substance in its solid state

Homework Equations



Q = m c Δθ

3. The Attempt at a Solution

Let mi = mass of the iron block; ms = mass of the substance; cs = specific heat of the substance; ci = specific heat of the iron; θe = the equilibrium temperature; θi = the initial temperature of the iron block; θs = the initial temperature of the substance

The process is thermally isolated, so Q received = Q given ⇒ [(ms)(cs)(θe - θs)] = [(mi)(ci)(θe-θi)] ⇒

mi/ms is constant = 0,8; θs = 20 and ci = 0,1 ⇒ (cs)(θe - θs) = 0,08(θe-θi) ⇔ cs = 0,08(θe-θi)/(θe - 20)


So, here's the problem. In the graph I initially picked the (150;50) point because it gives me the easy 100, so the cs = 4/15 cal/g ºC, but upon checking online the only answer I found (albeit not in a truly trustworthy source) is 0,28 cal/g ºC. Getting back to the exercise, this is the answer that we get when we input (200;60). (150;50) = 4/15 cal/g ºC; (100;40) = 0,24 cal/g ºC; (50;30) = 0,16 cal/g ºC.

My forced conclusion here is that the specific heat of the substance in its solid state is not constant (because the question did say the iron one was constant). So how should I know that the correct one was the (200;60)? Is my conclusion that cs is not a constant correct?

PS: Just to be sure, in the graph of the image, the Y axis are the equilibrium temperatures, the X - axis are the different temperatures of the iron block.

PS2: This is from an acceptance exam from 1997 and there are no definitive solutions of this online, so, again, I'm not entirely sure about the 0,28 cal/ g ºC

PS3: I wasn't able to host the image for some reason, in the preview it never showed up.

PS4: I'm sorry if I broke any rule, it's my first time posting.
 

Answers and Replies

  • #2
mjc123
Science Advisor
Homework Helper
1,077
522
You are not reading your graph carefully. It is small and unclear, I admit, but it is perfectly clear that the last point you mention is not (50,30) but about (50,26); i.e Te doesn't increase by 10° for every 50° increase in Ti, but a bit more.
 
  • Like
Likes frkCarl
  • #3
4
0
Damn, I can't believe I spent so much time on this because I simply wasn't careful reading the graph :cry::cry:. Upon further inspection in my printed graph, the only point which is exact from those I cited is (200;60), even (150;50) and (100;40) are slightly off.

Thank you so much for pointing this out, mjc123.
 
  • #4
20,857
4,544

Homework Statement


A researcher studies the heat transference between an iron block and an unknown substance, inside an adiabatic calorimeter. In successive experiences he puts inside the calorimeter the unknown substance, always in its solid state and with temperature θs = 20ºC, and the iron block, in each case with a different initial temperature "θi". In each case he measures the equilibrium temperature "θe". The results are shown in the graphic.

The quotient of the mass of the iron block and the mass of the unknown substance, "mi/ms", is equal to 0,8.
Consider the specific heat of the iron = 0.1 cal/g ºC

Image:
https://imgur.com/a/QLu3NQI

Question: Find the specific heat, "cs", of the unknown substance in its solid state

Homework Equations



Q = m c Δθ

3. The Attempt at a Solution

Let mi = mass of the iron block; ms = mass of the substance; cs = specific heat of the substance; ci = specific heat of the iron; θe = the equilibrium temperature; θi = the initial temperature of the iron block; θs = the initial temperature of the substance

The process is thermally isolated, so Q received = Q given ⇒ [(ms)(cs)(θe - θs)] = [(mi)(ci)(θe-θi)] ⇒
The above equation is incorrect. It should read:

[(ms)(cs)(θe - θs)] + [(mi)(ci)(θe-θi)] =0

or, equivalently,

Heat lost by iron = Heat gained by solid

where

heat lost by iron ##=m_ic_i(\theta_i-\theta_e)##

and

heat gained by solid ##=m_sc_s(\theta_e-\theta_s)##

So, the solution to the correct equation is $$\theta _e=\frac{m_sc_s}{(m_sc_s+m_ic_i)}\theta_s+\frac{m_ic_i}{(m_sc_s+m_ic_i)}\theta_i$$
So the slope is ##\frac{m_ic_i}{(m_sc_s+m_ic_i)}## and the intercept is ##\frac{m_sc_s}{(m_sc_s+m_ic_i)}\theta_s##

This leads to a value of cs of 0.28 cal/gm.C
 
Last edited:

Related Threads on Problem involving specific heat and graph

Replies
11
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
3
Views
337
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
7K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
1K
Top