Specific heat capacity and latent heat

  • Thread starter Thread starter waltssillyhat
  • Start date Start date
Click For Summary
The discussion revolves around calculating the final temperature of an ice cream mixture using specific heat capacity and latent heat concepts. The user initially attempted to equate the total heat extracted from the mixture to the product of mass, specific heat capacity, and temperature change (mcΔt) but encountered difficulties. They considered incorporating the brine solution's specific heat capacity but felt it complicated the problem. Ultimately, the correct final temperature was determined to be -1.6 degrees Celsius. The user expressed gratitude for the assistance received in solving the problem.
waltssillyhat
Messages
5
Reaction score
1
Homework Statement
this is a multi-part question from my textbook: "It is possible to make home-made ice cream by using a salt-water solution (brine) as a refrigerant for the cream and sugar mix. You can assume for the purposes of this exercise that the cream and sugar mix is 70% water. It is actually the water in the cream that freezes. Salt is added to ice to make 5.00 kg of cold brine at−11°C. 500 g of cream and sugar mixture is cooled in a plastic bag which is plunged into the refrigerant."

we have been given the following information:
Heat of fusion of water: 334 kJ kg−1
Specific heat capacity of water: 4200 J kg−1 K−1
Specific heat capacity of cream and sugar mix: 3.80 × 103 J kg−1 K−1
Specific heat capacity of brine solution: 3.5 × 103 J kg−1 K−1

i have found the first few parts of the question, which are the amount of heat extracted from the ice cream mixture to decrease its temperature to 0 degrees (47.5kJ) and the the heat that needs to be removed for the 70% of water in the mixture to freeze (116.9 kJ). the final part of the question asks "Calculate the final temperature of the mixture when all the water in the ice cream is frozen, assuming no other heat loss", which is where I am struggling.
Relevant Equations
Q = mcΔt and Q = m x Lfusion
I tried to set the total amount of heat extracted from the mixture (-116.9 + (-47.5) = -164.4 kJ) equal to mcΔt of the ice cream mixture and solve for the final temperature, but my answer was wrong. I'm just not really sure when to use the specific heat capacity of the brine solution. I was thinking of setting mcΔt of the brine solution and mcΔt of the ice cream mixture together and solving for the final temperature, but I would have to calculate the final temperature of the brine solution and I feel like this just complicates the question. By the way, the answer to the question is -1.6 degrees celsius. Any help is appreciated :)
 
Physics news on Phys.org
Create an unknown for the final temperature. Using that write expressions for:
- the total heat needs to be extracted from the ice cream mixture
- the total heat that will be absorbed by the brine
 
I was able to solve it thank you so much :)
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat (moth as source and bat as observer): $$f_2=\frac{333}{333-v}\times f_1$$ $$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$ Solving this equation, I get ##v=6.1## m/s but the answer key is...