How many grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C?
[mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute all of the numbers; mass cool H2O @ 20C is the only unknonwn.
so its setup like
(20*1.00)*(90-50)+(20*1.00)*(90-50) = 1600g?
To solve this problem, we can apply the principle of conservation of energy. We know that the amount of energy gained or lost by a substance is given by the equation:
Q = mcΔT
Where Q is the energy gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
In this case, we want to calculate the amount of energy that needs to be lost by the 800g of water at 90°C to reach 50°C. The specific heat capacity of water is approximately 4.18 J/g°C.
First, we calculate the energy lost by the 800g of water:
Q1 = mcΔT
Q1 = (800g)(4.18 J/g°C)(90°C - 50°C)
Q1 = (800g)(4.18 J/g°C)(40°C)
Q1 = 133,760 J
Next, we calculate the energy gained by the 20°C water:
Q2 = mcΔT
Q2 = mcΔT = (m)(4.18 J/g°C)(50°C - 20°C)
Since we know the energy gained by the 20°C water must be equal to the energy lost by the 800g of water, we can set up the equation:
Q1 = Q2
133,760 J = (m)(4.18 J/g°C)(50°C - 20°C)
Now, we can solve for the mass of the 20°C water:
(m)(4.18 J/g°C)(50°C - 20°C) = 133,760 J
(m)(4.18 J/g°C)(30°C) = 133,760 J
m = 133,760 J / [(4.18 J/g°C)(30°C)]
m ≈ 1,276.27 g
Therefore, approximately 1,276.27 grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C.