Question

what would be the final temperature of a mixture of 50 grams of water at 20 degree celsius temperature and 50 grams of water at 40 degree celsius temperature....

Answer 1

Hello Friend......

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The answer of u r question is......

Ans:

Mass(m1). =50g

Temperature(T1)=20°c

Mass(m2). =50g

Temperature(T2)=40°c

Specific heat (S)
of water =1cal/g-°c

Final temperature of the mixture,

T=[m1 T1+m2 T2]/m1+m2

=50×20+50×40/50+50

=1000+2000/100=3000/100

=30°c

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Thank you....⭐️⭐️⭐️⭐️

Answer 2

Solution -

According to the question we know that
M1 or M2 = 50g
T1 = 20°C , T2 = 40°C

$s = \frac{1cal}{g - c}$


Now, Temperature of the mixture is

$t = \frac{m1 \: t1 + m2 \: t2}{m1 \: + m2}$

$\frac{1000 + 2000}{100}$


$= \frac{3000}{100}$


= 30°c


I hope it's help you

Regards : Sangela