Could you help me with this thermodynamics question?

In summary: The heat required to increase the temperature of just the milk by 1 K comes from the Joules of energy that were added to the milk.
  • #1
Zeynaz
29
0
Homework Statement
A baby bottle contains 180mL milk. The heat capacity of the bottle and the milk is 770 J/K. The density of the milk is 1.04 kg/dm^3

a) calculate the heat capacity of the bottle
Relevant Equations
Q=C*dT
Q=cm*dT
Density =mass/Volume
So, I converted the V (milk) to m3 and found 1.8E-4 m3 and i already know the density so i found the mass of the milk in the bottle.

Mmilk= 1.9E-7 kg
Normally i would try to connect it with the formulas above but i don't know temperature. I am not sure how i can connect the dots.

Can someone help me about this?

Thanks!
 
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  • #2
Probably you can assume that the specific heat capacity for the milk is the same as for water.
How many Joules of heat must be added to the milk alone in order to raise the temperature of the milk by 1 Kelvin?

Also, check your calculation of the milk's mass.

EDIT: I don't think you should assume that the specific heat capacity of the milk is the same as for water. Were you given a value for milk, maybe in a table?
 
Last edited:
  • #3
TSny said:
Probably you can assume that the specific heat capacity for the milk is the same as for water.
How many Joules of heat must be added to the milk alone in order to raise the temperature of the milk by 1 Kelvin?

Also, check your calculation of the milk's mass.

EDIT: I don't think you should assume that the specific heat capacity of the milk is the same as for water. Were you given a value for milk, maybe in a table?
Yes you are right there is a table where i can find the density and specific heat capacity of all substances. So in this case the SHC of milk is 3.9E3.
And i think i found the right mass for the milk which is 0.187kg. The answer should be 40J/K but i don't know how to get there
 
  • #4
Zeynaz said:
Yes you are right there is a table where i can find the density and specific heat capacity of all substances. So in this case the SHC of milk is 3.9E3.
And i think i found the right mass for the milk which is 0.187kg.
So i know the combines SHC of milk and the bottle. I know the SHC of the milk. Am i just supposed take it as an average and say: SHC bottle+milk= (SCHmilk +SHCbottle)/2 ?
 
  • #5
Zeynaz said:
Am i just supposed take it as an average and say: SHC bottle+milk= (SCHmilk +SHCbottle)/2 ?
No, that won't work. Did you try to calculate the amount of heat required to increase the temperature of just the milk by 1 Kelvin?

How much heat would be required to increase the total milk-and-bottle system by 1 Kelvin?
 
  • #6
TSny said:
No, that won't work. Did you try to calculate the amount of heat required to increase the temperature of just the milk by 1 Kelvin?

How much heat would be required to increase the total milk-and-bottle system by 1 Kelvin?
For Q-milk= m,milk * c = 729.3 J
For Q Bottle-Milk= (m-milk+m-bottle)*770
But i don't know mass of the bottle.
Meanwhile, i also tried to find the mass of the bottle first by equating Q milk with Q bottle and i got 0.760kg and tried to work my way but it didnt really work.
Another way,
When i write: Qmilk-bottle = Q-bottle + Q milk and apply mass of the bottle, it doesn't give me the right answer and also i don't know Q b-m
 
  • #7
Zeynaz said:
Q-milk= m,milk * c = 729.3 J
OK. This is the heat that must be added to the milk alone in order to increase its temperature by 1 K.
For Q Bottle-Milk= (m-milk+m-bottle)*770
But i don't know mass of the bottle.
Note that the value of 770 J/K is not the specific heat capacity, ##c##, of the system. Rather, it is the heat capacity, ##C##, of the system. It is important to distinguish these. Can you express in words the meaning of the quantity 770 J/K?

Meanwhile, i also tried to find the mass of the bottle first by equating Q milk with Q bottle and i got 0.760kg and tried to work my way but it didnt really work.
There is no reason to expect that Q for the milk will be the same as Q for the bottle.
 
  • #8
TSny said:
OK. This is the heat that must be added to the milk alone in order to increase its temperature by 1 K.
Note that the value of 770 J/K is not the specific heat capacity, ##c##, of the system. Rather, it is the heat capacity, ##C##, of the system. It is important to distinguish these. Can you express in words the meaning of the quantity 770 J/K?

There is no reason to expect that Q for the milk will be the same as Q for the bottle.
That to increase 1 Kelvin of bottle-milk system we need 770 joules so in that case, technically i can take 770 as Q for bottle-milk system right?
 
  • #9
Yes. You also know the heat required to raise the temperature of just the milk by 1 K. Can you deduce how much heat is required to raise the temperature of just the bottle by 1 K?
 
  • #10
Thank you for your help I figured it out!
 
  • #11
OK, good work.
 

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the relationship between heat and other forms of energy, such as work. It also includes the study of how energy is transferred and transformed within a system.

2. Why is thermodynamics important?

Thermodynamics is important because it helps us understand how energy is used and transferred in various processes, such as in engines or chemical reactions. It also has practical applications in fields such as engineering, chemistry, and biology.

3. How do you solve a thermodynamics question?

To solve a thermodynamics question, you need to first identify the given variables, determine the type of thermodynamic process (e.g. isothermal, adiabatic, etc.), and then apply the appropriate equations and principles, such as the first and second laws of thermodynamics.

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Some common mistakes when solving thermodynamics problems include not correctly identifying the type of process, not considering all relevant variables, and not applying the correct equations or principles. It is also important to pay attention to units and ensure they are consistent throughout the calculation.

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To improve your understanding of thermodynamics, you can review the basic principles and equations, practice solving different types of problems, and seek help from a tutor or professor if needed. It can also be helpful to visualize thermodynamic processes and their effects on a system using diagrams or graphs.

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