Question

2 litres of water at 80 o C is found into a plastic bucket containing 10 litres of water at 20 o C. What is the final temperature of water (Density of water is 1kg/ litre) (neglect the heat gained by the bucketand specific heat capicity of water = 1 cal/ gm o C)

Answer 1

c = 1 cal / (g · °C) = 4,19 J / (g · °C) = 4,19 kJ / (kg · °C)
V₁ = 2 litres ⇒ m₁ = V₁d = 2 * 1 [kg/ltr * ltr = kg] = 2 kg
T₁ = 80 °C
m₂ = 10 kg
T₂ = 20 °C
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 = (2 · 80 + 10 · 20) ÷ (2 + 10) = 360 ÷ 12 = 30 [kg ·°C / kg = °C]

Answer: The final temperature is 30 °C.(The value of the specific heat capacity of water was not needed).