Question

What is the final temperature when 63.4 g of water at 80.3°C is mixed with 39.5 g of water at 26.1°C? (The specific heat of water is 4.184 J/g °C.) T- Joc

Answer 1

Answer : - 59.5 °C

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Given :-

1. 63.4 g water at temperature 80.3 °C

2. 39.5 g water at temperature 26.1 °C

The specific heat of water = C = 4.184 J/g. °C

Formula :-

  
q=mCAT
or  
-q=mCAT

i.e. ​​​​​​
q = m (T; -T;)
or

-9 = mc(T; -T;)
...(1)

where,

q = heat gain , - q = heat lost

m = mass

C = specific heat

Tf = final temperature   

Ti = initial temperature

When two water sample mix with each other

i) sample with high temperature (80.3 °C) lost heat and its temperature decrease

ii) sample with low temperature (26.1 °C) gain heat and its temperature increase

And we have,
-Host = 4gain
...(2)

First using equation (1) we can calculate, $q_{lost}$ and $q_{gain}$

Thus,

-Olost = mCT:-T;

i.e.
Glost = -mC(T-T;

i.e.  

910st = -63.4(g) x 4.184(J/g.°C)[T; -80.3(°C)]

And

Again = 39.5(g) x 4.184(J/g. C)[T: - 26.1(c)

Therefore from equation (2) we have

-63.4(g) x 4.184(J/g. C)[T; -80.3(°C)) = 39.5(g) x 4.184(J/g. C)[T;- 26.1(°C)

i.e.

–63.4(g) x [T; -80.3(°C)) = 39.5(g) x [T;- 26.1(C)
i.e.

-63.4(g)_[T; – 26.1(°C)] 39.5(g) T:- 80.3(°C)]

i.e.  

  
-1.61 = T: -26.1°C 1 [T; - 80.3(°C)]

i.e.

-1.61[T; - 80.3("C)] = [T; – 26.1(c)

i.e

  
-1.61T; +129.28(°C) =T; - 26.1(C)

i.e.

  

26.1(c) +129.28(°C) =T: +1.61T;

i.e.  

  
155,38(°C) = 2.61T;

i.e.   

T; = 59.53(C)

i.e.  
T; = 59.5(c)

(up to three significant figure)

Therefore, the final temperature when 63.4 g of water at 80.3 °C is mixed with 39.5 g of water at 26.1 °C is 59.5 °C

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