How many grams of steam at 100C must be mixed with 200g of water at 20C in order for the equilibrium temperature to be 50C?
To find the answer to this question, we can use the principle of conservation of energy.
The heat gained by the steam must be equal to the heat lost by the water. The formula to calculate heat transfer is:
Q = m * c * ΔT
where Q is the heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
For the steam:
Q₁ = m₁ * c₁ * ΔT₁
For the water:
Q₂ = m₂ * c₂ * ΔT₂
Since they reach equilibrium, Q₁ = Q₂. We can set them equal to each other:
m₁ * c₁ * ΔT₁ = m₂ * c₂ * ΔT₂
Rearranging the equation, we get:
m₁/m₂ = (c₂ * ΔT₂) / (c₁ * ΔT₁)
Given the values:
m₂ = 200g (mass of water)
c₂ = 4.18 J/g°C (specific heat capacity of water)
ΔT₂ = 50°C - 20°C = 30°C
We need the value of m₁ (mass of steam). To find it, we need the specific heat capacity of steam (c₁) and the change in temperature (ΔT₁).
Since steam is at its boiling point of 100°C, it will condense into water at 100°C when mixed with the water. Therefore, ΔT₁ = 100°C - 50°C = 50°C.
The specific heat capacity of steam (c₁) is 2.03 J/g°C.
Plugging in the values, we get:
m₁/200 = (4.18 * 30) / (2.03 * 50)
Simplifying the equation, we find:
m₁/200 = 0.624
To find the value of m₁, we can cross-multiply and solve for m₁:
m₁ = 200 * 0.624
m₁ ≈ 124.8 grams
Therefore, approximately 124.8 grams of steam at 100°C must be mixed with 200 grams of water at 20°C in order for the equilibrium temperature to be 50°C.
the sume of heats is zero...
Heat to heat water + heat or condense steam+heat to cool steam=0
200*cw*(50-20)+Hv*m + m*Cw*(50-100)=0
solve for mass m.