If 200 g of water at 80°C is mixed with 300 g of water at 20°C, what will be the

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Question: If 200 g of water at 80°C is mixed with 300 g of water at 20°C, what will be the final temperature of the mixture? (2019)

Options:

Correct Answer: 40°C

Exam Year: 2019

Solution:

Using the formula m1c1T1 + m2c2T2 = (m1 + m2)cT, we find T = (200*80 + 300*20) / (200 + 300) = 40°C.

If 200 g of water at 80°C is mixed with 300 g of water at 20°C, what will be the final temperature of the mixture? (2019)

If 200 g of water at 80°C is mixed with 300 g of water at 20°C, what will be the final temperature of the mixture? (2019)

Step 1: Identify the masses and temperatures of the two water samples. We have 200 g of water at 80°C (m1 = 200 g, T1 = 80°C) and 300 g of water at 20°C (m2 = 300 g, T2 = 20°C).
Step 2: Write down the formula for finding the final temperature (T) when mixing two substances: m1c1T1 + m2c2T2 = (m1 + m2)cT. Here, c (specific heat) is the same for both water samples, so we can simplify the equation.
Step 3: Substitute the known values into the formula. Since c cancels out, we can rewrite it as: 200 * 80 + 300 * 20 = (200 + 300) * T.
Step 4: Calculate the left side of the equation: 200 * 80 = 16000 and 300 * 20 = 6000. So, 16000 + 6000 = 22000.
Step 5: Now, the equation looks like this: 22000 = 500 * T.
Step 6: To find T, divide both sides by 500: T = 22000 / 500.
Step 7: Calculate the final temperature: T = 44°C.

Heat Transfer – The question tests the understanding of heat transfer and thermal equilibrium when mixing two bodies of water at different temperatures.
Specific Heat Capacity – It involves the concept of specific heat capacity, although in this case, the specific heat of water is constant and cancels out.
Weighted Average – The final temperature is calculated using a weighted average based on the masses and temperatures of the two water samples.