Water of mass 120g at 50°c is added to 200g of water at 1 0°c and the mixture is well stirred . calculate the temperature of the mixture .(neglect heat losses to the surrounding)
sum of heats gained is zero.
heatgainedhotwater+heatgainedbycoolwater=0
120*c*(Tf-50)+200*c*(Tf-10)=0
solve for Tf
pls wats the answer
Thank u
to learn
water of mass 120g
To calculate the temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the hot water must be equal to the heat gained by the cold water in order to achieve thermal equilibrium.
The equation that represents this principle is:
m1 * c1 * (Tf - T1) = m2 * c2 * (T2 - Tf)
Where:
m1 = mass of the hot water (120g),
m2 = mass of the cold water (200g),
c1 = specific heat capacity of water (4.18 J/g°C),
c2 = specific heat capacity of water (4.18 J/g°C),
Tf = final temperature of the mixture (unknown),
T1 = initial temperature of the hot water (50°C),
T2 = initial temperature of the cold water (10°C).
Substituting the values into the equation:
(120g) * (4.18 J/g°C) * (Tf - 50°C) = (200g) * (4.18 J/g°C) * (10°C - Tf)
Now, let's solve the equation to find the value of Tf:
(120g) * (4.18 J/g°C) * Tf - (120g) * (4.18 J/g°C) * 50°C = (200g) * (4.18 J/g°C) * 10°C - (200g) * (4.18 J/g°C) * Tf
(120g) * (4.18 J/g°C) * Tf + (200g) * (4.18 J/g°C) * Tf = (200g) * (4.18 J/g°C) * 10°C + (120g) * (4.18 J/g°C) * 50°C
(120g + 200g) * (4.18 J/g°C) * Tf = (200g) * (4.18 J/g°C) * 10°C + (120g) * (4.18 J/g°C) * 50°C
(320g) * (4.18 J/g°C) * Tf = (200g) * (4.18 J/g°C) * 10°C + (120g) * (4.18 J/g°C) * 50°C
Simplifying the equation further:
(320g) * (4.18 J/g°C) * Tf = (200g) * (4.18 J/g°C) * 10°C + (120g) * (4.18 J/g°C) * 50°C
This equation can be solved to find the value of Tf, which will be the temperature of the mixture.