how would i calculate; a 50g sample of water at 100 degrees is poured into a 50g sample of water at 25 degrees. with a specific heat of water (l) being 4.184 j/g degrees C.what will be the final temperature of the water?

heat gained + heat lost = 0
[massH2O*specificheat*(Tf-Ti)+[massH2O*specific heat*(Tf-Ti)].
Tf is final T. Solve for this.
Ti is initial T of each.
Post your work if you get stuck and need further assistance. There is a short cut way of doing this but the above equations works for almost anything.

50g*4.184J/g*C*(Tf-100)+50g*4.184J/g*C*(Tf-25)=0

50g*4.184J/g*C*Tf-4184J-50g*4.184J/g*C*Tf+1046J=0

-3138J=0

Tf=3138J/50g*4.184J/g*C

Tf=37.7°C

To calculate the final temperature of the water after pouring the two samples together, you need to use the formula:

Heat gained + Heat lost = 0

The heat gained by the cooler water sample can be calculated using the equation: mass × specific heat × (final temperature - initial temperature). The heat lost by the hotter water sample can be calculated using the same equation.

In this case, the mass of both water samples is 50g, the specific heat of water (l) is 4.184 J/g°C, the initial temperature of the cooler water is 25°C, and the initial temperature of the hotter water is 100°C. The final temperature is denoted as Tf.

So the equation becomes:
50g × 4.184 J/g°C × (Tf - 25°C) + 50g × 4.184 J/g°C × (Tf - 100°C) = 0

Now, you can solve this equation for Tf. Start by expanding and simplifying:

50g × 4.184 J/g°C × Tf - 50g × 4.184 J/g°C × 25°C + 50g × 4.184 J/g°C × Tf - 50g × 4.184 J/g°C × 100°C = 0

Combine like terms:

2 × 50g × 4.184 J/g°C × Tf - 50g × 4.184 J/g°C × 25°C - 50g × 4.184 J/g°C × 100°C = 0

100g × 4.184 J/g°C × Tf - 50g × 4.184 J/g°C × 25°C - 50g × 4.184 J/g°C × 100°C = 0

Factor out the common term:

100g × 4.184 J/g°C × Tf - 50g × 4.184 J/g°C (25°C + 100°C) = 0

100g × 4.184 J/g°C × Tf - 50g × 4.184 J/g°C × 125°C = 0

Divide both sides by the coefficient of Tf:

100g × 4.184 J/g°C × Tf / (50g × 4.184 J/g°C) = 125°C

Cancel out the units and simplify:

Tf = 125°C

Therefore, the final temperature of the water will be 125°C.

To calculate the final temperature of the water, you can apply the principle of heat gained and heat lost being equal.

1. Calculate the heat gained by the 50g sample of water at 100 degrees Celsius:
Heat gained = mass * specific heat * temperature change
Heat gained = 50g * 4.184 J/g°C * (final temperature - 100°C)

2. Calculate the heat lost by the 50g sample of water at 25 degrees Celsius:
Heat lost = mass * specific heat * temperature change
Heat lost = 50g * 4.184 J/g°C * (final temperature - 25°C)

3. Set up the equation:
Heat gained + Heat lost = 0
50g * 4.184 J/g°C * (final temperature - 100°C) + 50g * 4.184 J/g°C * (final temperature - 25°C) = 0

4. Simplify the equation and solve for the final temperature:
50g * 4.184 J/g°C * (final temperature - 100°C) + 50g * 4.184 J/g°C * (final temperature - 25°C) = 0
208.4g°C * (final temperature - 100°C) + 208.4g°C * (final temperature - 25°C) = 0
208.4g°C * final temperature - 20840 J + 208.4g°C * final temperature - 5210 J = 0
416.8g°C * final temperature - 26050 J = 0
416.8g°C * final temperature = 26050 J
final temperature = 26050 J / 416.8g°C

5. Calculate the final temperature:
final temperature = 62.5°C

Therefore, the final temperature of the water will be 62.5 degrees Celsius.