If 200 g of water at 80°C is mixed with 300 g of water at 20°C, what will be the
Practice Questions
Q1
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? (Assume no heat loss to the surroundings)
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Questions & Step-by-Step Solutions
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? (Assume no heat loss to the surroundings)
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: Recognize that the specific heat capacity (c) of water is the same for both samples, so we can use 'c' as a common factor in our calculations.
Step 3: Write down the heat transfer equation: m1 * c * T1 + m2 * c * T2 = (m1 + m2) * c * Tfinal.
Step 4: Since 'c' is the same for both samples, we can cancel it out from the equation, simplifying it to: m1 * T1 + m2 * T2 = (m1 + m2) * Tfinal.
Step 5: Substitute the values into the equation: 200 g * 80°C + 300 g * 20°C = (200 g + 300 g) * Tfinal.
Step 6: Calculate the left side: 16000 + 6000 = 22000.
Step 7: The equation now looks like: 22000 = 500 g * Tfinal.
Step 8: Solve for Tfinal by dividing both sides by 500 g: Tfinal = 22000 / 500.
Step 9: Calculate Tfinal: Tfinal = 44°C.
Heat Transfer – The principle of conservation of energy applied to thermal systems, where heat lost by the hotter substance equals heat gained by the cooler substance.
Specific Heat Capacity – The amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius.
Equilibrium Temperature – The final temperature reached when two bodies at different temperatures are mixed, assuming no heat loss to the environment.
