I can't believe it my Refrigerator is broken, Help!

Classical Mechanics Level 4

Imagine that on this particular hot day, you want to prepare a cold glass of refreshing lemonade to cool yourself down. Upon pouring $200cm^{ 3 }$ of lemonade into your cup, you realized that the freezer was broken! All the remaining ice had already melted and there was nothing left. Another problem was that the sun's heat was getting more intense and the lemonade that you prepared went up to $70^{ 0 }C$ . Even by drinking it at such a temperature, it would not be effective in cooling you down in this scorching heat.

As you walked frantically around the living room thinking of a suitable idea, the doorbell suddenly rang. When you opened the door, it was Joseph Black (picture above). He was a scientist and knew the situation that you are going through(Though he didn't tell you how he knew it, it's creepy). However, Joseph wasn't that too generous, he said if you could determine accurately the exact mass of ice needed to cool it down, not only will Joseph grant you a quick repair of his refrigerator ,but also a complementary gift of $100g$ of ice cubes, if not, you would have to continue suffering in the heat. Now, here's what Joseph wants you to answer:"Find the mass of ice required to decrease the temperature of the lemonade to $25^{ 0 }C$ ; so, did you figure it out the way to go about this or are you prepared to be scorched?"

Details :

Assume that the specific heat of fusion of ice is $3.40\times { 10 }^{ 5 }J{ kg }^{ -1 }$ and that the specific heat capacity of water is to be $4200J{ kg }^{ -1 }{ K }^{ -1 }$ .
Assume that the lemonade drink is mainly made up of water.
Input your answer in $g$ and in $3s.f.$ .
In case if you didn't know, Joseph Black was a true pioneer in discovering latent heat. In fact, in the 1800s, he deduced that the application of heat to ice at its melting point does not cause a rise in temperature of the ice/water mixture, but rather an increase in the amount of water in the mixture. In addition, Black observed that the application of heat to boiling water does not result in a change in temperature of the water/steam mixture, but rather an increase in the amount of steam. From these observations, he could conclude that the heat applied must have combined with the ice particles and boiling water and become latent. Thus, Joseph coined the term, " latent heat ". For more information about his discovery, go to Joseph Black research .

The answer is 84.9.

1 solution

Pranjal Jain
Jan 5, 2015

Mass of lemonade (assuming its density is equal to that of water) $=200\text{cm}^{3}×1\text{g cm}^{-3}=200g=0.2kg$

Heat released by lemonade to cool down from $70°C$ to $25°C=0.2×(70-25)×4200=37800 J$

For the sake of energy conservation, $37800 J$ energy must be absorbed by ice.

Now assume the required mass of ice be $x$ kg

Heat absorbed by ice due to melting $=x×3.4×10^5$

Heat absorbed by ice for increasing temperature from $0°C$ to $25°C=x×(25-0)×4200=x×1.05×10^5$

Total heat absorbed $=4.45×10^5×x=37800\\\Rightarrow x=\frac{37800}{4.45×10^5}=0.0849$

So the mass of ice is $0.0849$ kg or $84.9$ grams.

I input wrong answer on my first two attempts by answering in Kg instead of g ! :(

Snehal Shekatkar - 6 years, 5 months ago

From heat balance. Quantity of heat reject by ice = Quantity of heat recived by warm water. Mass of lce (in gram) s.f of ice in (j/gram)=mass of warm water (in gram) heat capcity of water(j/gram.k) temprature differance of water. 3 10^5 mass of ice=200 4200*(70-25). mass of ice = 126 gram.

محمد فكرى - 6 years, 5 months ago

Since final temperature of system is 25°C, ice will release heat when taken from 0°C to 25°C.

Pranjal Jain - 6 years, 5 months ago

thats what i got, but i typed it in as kilograms. ugh.

Daniel Golden - 6 years, 5 months ago

I can't believe it my Refrigerator is broken, Help!

The answer is 84.9.

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