Question

the
would be
final temperature of a
3. what
mixture of Gog
of water at 40°C temperature and
Goog
of water at 8ic temperature a​

Answer 1

Correct $\huge\mathfrak\red{Question}$

What would be the final temperature of of mixture if 40 grams of water at 20 degrees celsius mixed with 60 grams of of water at 30 degrees Celsius.

$\huge\mathfrak\red{Answer}$

since it is given that 40 gms water at 20 ⁰ C and 40 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.

=====================

final temperature of the mixture =

= [ m1 * T1 + m2 * T2 ] / (m1 + m2)

= [ 40 gms * 20⁰ C + 40 gms * 40⁰C ] / (40+40)

= 800 / 100 = 80⁰C

====================

another way using specific heats :

let the final temperature be = T ⁰C

Amount of heat given out by the hot water = m * s * (40⁰C - T)

= 50 gms * s* (40 -T)

Amount of heat taken in by the cold water = m * s * (T - 20⁰C)

= 50 gms * s * (T - 20 )

As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,

50 * s * (40 -T) = 50 gm * s * (T-20)

40 - T = T - 20

2 T = 60 => T = 30⁰C

Answer 2

Question

What would be the final temperature of of mixture if 40 grams of water at 20 degrees celsius mixed with 60 grams of of water at 30 degrees Celsius.

Answer

since it is given that 40 gms water at 20 ⁰ C and 40 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.

=====================

final temperature of the mixture =

= [ m1 * T1 + m2 * T2 ] / (m1 + m2)

= [ 40 gms * 20⁰ C + 40 gms * 40⁰C ] / (40+40)

= 800 / 100 = 80⁰C

====================

another way using specific heats :

let the final temperature be = T ⁰C

Amount of heat given out by the hot water = m * s * (40⁰C - T)

= 50 gms * s* (40 -T)

Amount of heat taken in by the cold water = m * s * (T - 20⁰C)

= 50 gms * s * (T - 20 )

As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,

50 * s * (40 -T) = 50 gm * s * (T-20)

40 - T = T - 20

2 T = 60 => T = 30⁰C

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