A student mixes 400 grams of cold water with a temperature of 5 degrees Celsius with tea with a mass of 800 grams at a temperature of 40 degrees Celsius. (Liquid = 2100 j / kg celsius and c water = 4200 j / kg celsius)
Answers
Discussion
For questions like this, we use the Black Principle formula.
Black Principle
Q enter = Q out
A. Final mixture temperature (t1)
Known :
• m1 = 400 g
• m2 = 800 g
• t water = 5 °
• t tea = 40 ° c
• C2 = 2100 j / kg
• C1 = 4200 j / kg
Settlement:
Q enter = Q out
m1. c1. (t1 - t air) = m2. c2. (t tea - t1)
400. 4200. (t1 - 5) = 800. 2100. (40 - t1)
t1 - 5 = 800/400. 2100/4200. (40 - t1)
t1 - 5 = 2. 1/2. (40 - t1)
t1 - 5 = 40 - t1
t1 + t1 = 40 + 5
2t1 = 45
t1 = 45/2
t1 = 22.5 ° C
So, the final temperature of the mixture between water and tea is 22.5 ° C
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B. Lots of mixed water (m1)
Known :
m2 = 800 g
t water = 5 °
t tea = 40 ° c
C1 = 2100 j / kg
C2 = 4200 j / kg
t1 = 20 ° c
Settlement:
Q enter = Q out
m1. c1. (t1 - t air) = m2. c2. (t tea - t1)
m1. 4200. (20 - 5) = 800. 2100. (40-20)
m1. 15 = 800. (2100/4200). 20
m1 = 800. 1/2. 20/15
m1 = 400. 4/3
m1 = 1600/3
m1 = 533.33 g
So, a lot of water is needed so that the temperature to be 20 ° C is 2133.33 grams.
The enthalpy of fusion of ice is 334 J/g. The specific heat of water is 4.2 J/g.
To cool 50 g of water from 50 °C to 0 °C would require the removal of
4.2 x 50 x 50 =10500 J.
To melt the ice would require the addition of
334 x 10 = 3340 J
10500 > 3340 thus you can melt all the ice and have some heat to spare, specifically 10500 - 3340 = 7160 J
Now use this to warm up 10 + 50 = 60 g of water at 0 °C
7160 / (60 x 4.2) = 28.4 °C
You will get a slightly different number if you use 4.184 instead of 4.2 J/g.
Hope this explains it to you rather than just doing your homework.