10.2.3. Example of Heat Calculation

To heat 2 kg of water from 20°C to 100°C, with a specific heat capacity of 4.18 kJ/kg°C, the heat required is:

Q=2×4.18×(100−20)=2×4.18×80=669.6 J Q = 2 \times 4.18 \times (100 - 20) = 2 \times 4.18 \times 80 = 669.6 , \text{J} Hence, 669.6 Joules of heat is required to heat the water.

This calculation demonstrates how to find the amount of heat needed to raise the temperature of water. Here, we're trying to heat 2 kilograms of water from an initial temperature of 20°C to a final temperature of 100°C.

1. Specific Heat Capacity: The specific heat capacity of water is given as 4.18 kJ/kg°C. This value represents the amount of heat needed to raise the temperature of one kilogram of water by one degree Celsius.
2. Formula: We use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
3. Calculate ΔT: First, calculate the temperature change (ΔT): 100°C - 20°C = 80°C.
4. Plugging Values: Now plug the values into the formula: Q = 2 kg * 4.18 kJ/kg°C * 80°C.
5. Final Calculation: This means Q = 2 × 4.18 × 80 = 669.6 J. This is the total amount of heat required to heat the water.

So, the result shows that 669.6 Joules of heat energy needs to be added to the water to achieve this temperature increase.

Imagine heating a large pot of water to make pasta. You start with cold water at room temperature and then turn the stove on. The heat from the stove transfers to the water, raising its temperature. The pot has to add enough heat energy to not only change the water's temperature but also to keep it at a rolling boil to cook properly. Similarly, this calculation helps understand the precise amount of energy required to make such a temperature change.

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