If 1 kg of water is heated from 25°C to 75°C, how much heat is absorbed? (Specif

If 1 kg of water is heated from 25°C to 75°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C) (2021)

If 1 kg of water is heated from 25°C to 75°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C) (2021)

Step 1: Identify the mass of water. Here, the mass (m) is 1 kg, which is equal to 1000 grams.
Step 2: Identify the specific heat of water. The specific heat (c) is given as 4.2 J/g°C.
Step 3: Determine the change in temperature (ΔT). The initial temperature is 25°C and the final temperature is 75°C. So, ΔT = 75°C - 25°C = 50°C.
Step 4: Use the formula for heat absorbed (Q). The formula is Q = mcΔT.
Step 5: Substitute the values into the formula: Q = (1000 g)(4.2 J/g°C)(50°C).
Step 6: Calculate the result: Q = 1000 * 4.2 * 50 = 210000 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 = mcΔT to calculate the heat absorbed or released during a temperature change.