please show me a simple formula to solve this
find the final equilibirum temperature when 10.0 g of milk at 10.0degC is added to 1.60 * 10^2 g of coffee with a temperature of 90.0degC. assume the specific heats of coffee and milk are the same as for water (Cp,w=4.19J/g*degC), and disregard the heat capacity of the container
thanks!
The sum of the heats gained is zero.
Heat gained = mass*c*(Tf-Ti)
So here...
massmilk*cmilk*(Tf-10)+masscoffee(ccoffee)(Tf-90)=0
If the c are the same, they divide out.
Solve for Tf
ok thnx
i'll try it
To solve this problem, we can use the principle of heat transfer. The formula to find the final equilibrium temperature is:
m1 * Cp1 * (Tf - Ti1) = m2 * Cp2 * (Tf - Ti2)
Where:
m1 is the mass of the first substance (milk) in grams
Cp1 is the specific heat capacity of the first substance (milk)
Ti1 is the initial temperature of the first substance (milk)
Tf is the final equilibrium temperature
m2 is the mass of the second substance (coffee) in grams
Cp2 is the specific heat capacity of the second substance (coffee)
Ti2 is the initial temperature of the second substance (coffee)
Given values:
milk: m1 = 10.0 g
coffee: m2 = 1.60 * 10^2 g
Specific heat capacity: Cp1 = Cp2 = 4.19 J/g°C (for water)
Initial temperatures: Ti1 = 10.0°C (milk), Ti2 = 90.0°C (coffee)
Substitute these values into the formula and solve for Tf:
10.0 * 4.19 * (Tf - 10.0) = 1.60 * 10^2 * 4.19 * (Tf - 90.0)
Now, you can solve the equation for Tf by distributing and simplifying:
41.9 * Tf - 419 = 670.4 * Tf - 60336
Combine like terms:
41.9 * Tf - 670.4 * Tf = -60336 + 419
Simplify further:
-628.5 * Tf = -59917
Divide both sides by -628.5:
Tf = -59917 / -628.5
Tf ≈ 95.3°C
Therefore, the final equilibrium temperature is approximately 95.3°C.