Hot water at a temperature of t is added to twice that amount of water at a temperature of 30°C. If the resulting temperature of the mixture is 50°C. Calcul...

Hot water at a temperature of t is added to twice that amount of water at a temperature of 30°C. If the resulting temperature of the mixture is 50°C. Calculate t.

Answer Details

To solve this problem, we need to apply the principle of mixtures. The principle states that the total heat content of a mixture of two or more substances is equal to the sum of their individual heat contents before mixing. Let's assume that the initial amount of hot water added is x. Therefore, the initial amount of cold water added is 2x. The heat content of the hot water is given by Q₁ = mxΔt₁, where m is the mass of the hot water, x is the amount added, and Δt₁ is the change in temperature of the hot water. Similarly, the heat content of the cold water is given by Q₂ = mcΔt₂, where m is the mass of the cold water, 2x is the amount added, and Δt₂ is the change in temperature of the cold water. Since the total heat content of the mixture is equal to the sum of the individual heat contents, we can write: mxΔt₁ + mcΔt₂ = (m+2m)cΔt where Δt is the change in temperature of the mixture, and we have used the fact that the mass of the hot water and cold water together is 3m. We know that the final temperature of the mixture is 50^oC, and the initial temperature of the cold water is 30^oC. Therefore, Δt = 50 - 30 = 20^oC. Simplifying the equation above, we get: xΔt₁ + 2xcΔt₂ = 3mcΔt Substituting the given values, we get: x(t - 100) + 2x(50 - 30) = 3(2x)(50 - 30) Simplifying this expression, we get: xt - 100x + 40x = 120x Solving for x, we get: x = 4t/3 Substituting this value in the expression for Δt₁, we get: Δt₁ = t - 100 = (3/4)x - 100 = (3/4)(4t/3) - 100 = t/3 - 100 Therefore, the initial temperature of the hot water is t/3 - 100. Now we can use the fact that the final temperature of the mixture is 50^oC to write: mxΔt₁ + mcΔt₂ = (m+2m)cΔt Substituting the values, we get: x(t/3 - 100 - 50) + 2x(50 - 30) = 3mx(50 - 30) Simplifying this expression, we get: xt/3 - 150x + 40x = 60mx Solving for t, we get: t = 90^oC Therefore, the initial temperature of the hot water is 90/3 - 100 = -70^oC. However, this does not make physical sense since the temperature of water cannot be negative. We made an error in our calculation