(phyics) question: calculate the final temperature of the water if 400grams of water at 80degress Celsius is poured into a 100grams pot at 30dgrees Celsius .

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82.64

To calculate the final temperature of the water, we can use the principle of heat transfer, which states that the heat gained by one object should be equal to the heat lost by the other object in a closed system.

We can use the equation:

(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0

where:
m1 = mass of water = 400 grams
c1 = specific heat capacity of water = 4.18 J/g°C
ΔT1 = change in temperature for water = final temperature - initial temperature of water
m2 = mass of pot = 100 grams
c2 = specific heat capacity of the pot (assuming it is made of a material with a specific heat capacity similar to water) = 4.18 J/g°C
ΔT2 = change in temperature for the pot = final temperature - initial temperature of the pot

Since we want to find the final temperature, we can rearrange the equation as:

(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0

(m1 * c1 * ΔT1) = - (m2 * c2 * ΔT2)

ΔT1 = - [(m2 * c2 * ΔT2) / (m1 * c1)]

Now, substitute the given values into the equation:

ΔT1 = - [(100 g * 4.18 J/g°C * (Tfinal - 30°C)) / (400 g * 4.18 J/g°C)]

Simplify the equation:

ΔT1 = - [(1 * (Tfinal - 30)) / 4]

Now, solve for ΔT1:

(Tfinal - 30) / 4 = ΔT1

Now, let's substitute ΔT1 back into the equation:

(Tfinal - 30) / 4 = - [(100 g * 4.18 J/g°C * ΔT2) / (400 g * 4.18 J/g°C)]

Simplify the equation further:

(Tfinal - 30) / 4 = - (ΔT2 / 4)

To cancel out the fractions, we can eliminate the denominators:

4(Tfinal - 30) = -ΔT2

Expand the equation:

4Tfinal - 120 = -ΔT2

Rearrange the equation to isolate Tfinal:

4Tfinal = -ΔT2 + 120

Tfinal = (-ΔT2 + 120) / 4

Now, substitute the given values into the equation:

Tfinal = (-(100 g * 4.18 J/g°C * ΔT2) / (400 g * 4.18 J/g°C)) + 120 / 4

Simplify the equation:

Tfinal = (- (100 g * ΔT2) / (400 g)) + 30

Now, calculate the change in temperature for the pot. Since the initial temperature of the pot is 30°C and the final temperature of the pot is the same as the final temperature of the water, we have:

ΔT2 = Tfinal - 30

Substitute this value back into the equation for Tfinal:

Tfinal = (- (100 g * (Tfinal - 30)) / (400 g)) + 30

Now, solve for Tfinal:

Multiplying through by 400g:

400g * Tfinal = - (100g * (Tfinal - 30)) + 12000g

Expanding and simplifying the equation:

400g * Tfinal = -100g * Tfinal + 3000g + 12000g

Combining like terms:

500g * Tfinal = 15000g

Divide both sides by 500g:

Tfinal = 15000g / 500g

Final temperature, Tfinal = 30°C

Therefore, the final temperature of the water and the pot will be 30 degrees Celsius.