Math, asked by NIGHTSANDSONIC, 1 year ago

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

Answered by NailanIca
3

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

••••••••••••••••••••••

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.

Answered by ans81
0
HEY MATE HERE IS YOUR ANSWER



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.

Similar questions