If 1 kg of water is heated from 20°C to 100°C, how much heat is absorbed? (Speci

₹0.0

Question: If 1 kg of water is heated from 20°C to 100°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C)

Options:

Correct Answer: 4200 J

Solution:

Q = m*c*ΔT = 1000 g * 4.2 J/g°C * (100°C - 20°C) = 4200 J.

If 1 kg of water is heated from 20°C to 100°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C)

If 1 kg of water is heated from 20°C to 100°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C)

Step 1: Identify the mass of water. Here, the mass (m) is 1 kg, which is equal to 1000 grams (since 1 kg = 1000 g).
Step 2: Identify the specific heat of water. The specific heat (c) is given as 4.2 J/g°C.
Step 3: Determine the initial and final temperatures. The initial temperature is 20°C and the final temperature is 100°C.
Step 4: Calculate the change in temperature (ΔT). ΔT = final temperature - initial temperature = 100°C - 20°C = 80°C.
Step 5: Use the formula for heat absorbed (Q). Q = m * c * ΔT.
Step 6: Substitute the values into the formula: Q = 1000 g * 4.2 J/g°C * 80°C.
Step 7: Calculate the result: Q = 1000 * 4.2 * 80 = 336000 J.
Step 8: Therefore, the total heat absorbed is 336000 J.

Specific Heat Capacity – The amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.
Heat Transfer Calculation – Using the formula Q = m*c*ΔT to calculate the heat absorbed or released during a temperature change.
Unit Conversion – Understanding the conversion between kilograms and grams, as well as the implications for calculations.