500g of water at 80 c is mixed with 250g of water at 50c . calculate the final temperature of the mixture?
Answers
T₁ = 80 °C
m₂ = 0.25 kg
T₂ = 50 °C
c = 4,19 kJ / (kg · °C) - heat capacity of water
Tₓ = final temperature of water: T₂ < Tₓ < T₁
On the basis of the laws of thermodynamics heat cast equals heat fetched:
H₁ = H₂
H₁ = m₁c(T₁ - Tₓ)
H₂ = m₂c(Tₓ - T₂)m₁c(T₁ - Tₓ) = m₂c(Tₓ - T₂)
m₁(T₁ - Tₓ) = m₂(Tₓ - T₂)
m₁T₁ - m₁Tₓ = m₂Tₓ - m₂T₂
m₂Tₓ + m₁Tₓ = m₁T₁ + m₂T₂
Tₓ = (m₁T₁ + m₂T₂) ÷ (m₁ + m₂)
Tₓ = (0.5 · 80 + 0.25 · 50) ÷ (0.5 + 0.25) = 52.5 ÷ 0.75 = 70 [kg ·°C / kg = °C]
Answer: The final temperature is 70 °C.(The value of the specific heat capacity of water was not needed).
Answer:
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 = 30⁰
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