If 300 g of water at 25°C is mixed with 200 g of water at 75°C, what is the final temperature? (Specific heat of water = 4.2 J/g°C)

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Q: If 300 g of water at 25°C is mixed with 200 g of water at 75°C, what is the final temperature? (Specific heat of water = 4.2 J/g°C)

Solution: Using m1*T1 + m2*T2 = (m1 + m2)*Tf, we find Tf = (300*25 + 200*75) / (300 + 200) = 50°C.

Steps: 8

Step 1: Identify the masses and temperatures of the two water samples. We have 300 g of water at 25°C (m1 = 300 g, T1 = 25°C) and 200 g of water at 75°C (m2 = 200 g, T2 = 75°C).
Step 2: Write down the formula to find the final temperature (Tf) when two masses of water are mixed: m1*T1 + m2*T2 = (m1 + m2)*Tf.
Step 3: Substitute the values into the formula: 300*25 + 200*75 = (300 + 200)*Tf.
Step 4: Calculate the left side: 300*25 = 7500 and 200*75 = 15000. So, 7500 + 15000 = 22500.
Step 5: Calculate the total mass: 300 + 200 = 500 g.
Step 6: Now we have the equation: 22500 = 500*Tf.
Step 7: To find Tf, divide both sides by 500: Tf = 22500 / 500.
Step 8: Calculate Tf: 22500 / 500 = 45°C.