How do you solve this equation? A piece of aluminum with a mass of 3.90g and a temperature of 99.3 degrees celcius is dropped into 10.0 g of a water at a temperature of 22.6 degrees celcius. What is the final temperature of the system?

(3.9)*(0.9025J)*(Tfinal-99.3)+(10)*(4.184J)*(Tfinal-22.6)=0

(3.51975J)*(Tfinal-99.3)+(41.84J)*(Tfinal-22.6)=0

45.35975JTfinal-1295.095175=0

1295.095175/45.35975=28.6 degrees celsius

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[masss Al x specific heat Al x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0

Solve for Tfinal.

To solve this equation, we can use the principle of heat transfer, specifically the equation for heat exchange between two objects:

q1 = q2

where q1 is the heat gained by the water, and q2 is the heat lost by the aluminum.

We can calculate the heat gained by the water using the formula:

q1 = m1 * c1 * ΔT1

where m1 is the mass of water, c1 is the specific heat capacity of water, and ΔT1 is the change in temperature of the water.

Similarly, we can calculate the heat lost by the aluminum using the formula:

q2 = m2 * c2 * ΔT2

where m2 is the mass of aluminum, c2 is the specific heat capacity of aluminum, and ΔT2 is the change in temperature of the aluminum.

Since the heat gained by the water is equal to the heat lost by the aluminum, we can set up the equation:

m1 * c1 * ΔT1 = m2 * c2 * ΔT2

Plugging in the given values:

m1 = 10.0 g (mass of water)
c1 = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = final temperature - initial temperature = final temperature - 22.6°C (change in temperature of water)

m2 = 3.90 g (mass of aluminum)
c2 = 0.897 J/g°C (specific heat capacity of aluminum)
ΔT2 = 99.3°C - final temperature (change in temperature of aluminum)

Now, we can substitute these values back into the equation:

10.0 g * 4.18 J/g°C * (final temperature - 22.6°C) = 3.90 g * 0.897 J/g°C * (99.3°C - final temperature)

Solving this equation will give us the final temperature of the system.