50 grams of water at 20 degree Celsius is mixed with 50 grams of water at 40 degree Celsius. what would be the final temperature of the mixture
Answers
Explanation:
since it is given that 50 gms water at 20 ⁰ C and 50 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.
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final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
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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:
30°C
Explanation :
Given,
Mass m1 = 50g
Temperature T1 = 40°C
Mass m2 = 50g
Temperature T2 = 20°C
By using method of mixtures,
Final temperature
T = m1T1 + m2T2/m1 +m2
= 50×40+50×20/50+50
= 2000+1000/100
= 30°C
The final temperature = 30°C