Heat Removal from Water: Calculation and Applications

Understanding how much heat must be removed from water to bring its temperature from one level to another is crucial in various scientific and real-world applications. This article delves into the calculation of heat removal using specific heat capacity, provides relevant formulas, and offers practical examples to enhance your comprehension.

Introduction to Specific Heat Capacity

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of one kilogram of the substance by one degree Celsius (or Kelvin). For liquid water, the specific heat capacity is approximately 4186 J/kg°C (4.186 kJ/kg°C).

Calculating Heat Removed from Water

The formula to calculate the heat removed from a substance is:

Q mcΔT

Where:

Q is the heat removed, measured in joules (J) m is the mass of the substance, measured in kilograms (kg) c is the specific heat capacity of the substance, measured in J/kg°C ΔT is the change in temperature, measured in degrees Celsius (°C)

Example Calculation

Let's consider the scenario of removing heat from 0.175 kg of water, bringing its temperature from 24.3°C to 5.00°C.

Calculate the change in temperature (ΔT): Substitute the values into the formula: Calculate the heat removed (Q):

1. Calculate the change in temperature:

[ Delta T T_{text{final}} - T_{text{initial}} 5.00, text{°C} - 24.3, text{°C} -19.3, text{°C}]

2. Substitute the values into the formula:

[ Q 0.175, text{kg} times 4186, text{J/kg°C} times (-19.3, text{°C})]

3. Calculate the heat removed (Q):

[ Q approx 0.175 times 4186 times -19.3 approx 14157.63, text{J}]

Since we are interested in the heat removed, we take the absolute value:

[ Q approx 14158, text{J}]

Thus, approximately 14158 joules of heat must be removed from the water to lower its temperature from 24.3°C to 5.00°C.

Additional Examples and Conversion

Let's explore additional calculations for different scenarios:

Using specific heat constant for liquid water (4.184 kJ/kg°C): Using a different specific heat value (4200 J/kg°C): Using mass in grams and specific heat in J/g°C:

1. Using the specific heat constant for liquid water (4.184 kJ/kg°C):

[ 0.175, text{kg} times 4184, text{J/kg°C} times 19.3, text{°C} 14185.5, text{J}]

2. Using a different specific heat value (4200 J/kg°C):

[ Q 0.175, text{kg} times 4200, text{J/kg°C} times 19.3, text{°C} approx 14185.5, text{J}]

3. Using mass in grams (175 g) and specific heat in J/g°C:

[ 175, text{g} times 4.18, text{J/g°C} times 19.3, text{°C} 14118, text{J}]

Conclusion

In conclusion, the specific heat capacity is a fundamental concept in thermodynamics, and understanding how to calculate heat removal is essential for tackling real-world problems. Whether you are dealing with small masses or larger ones, the formulas remain the same, providing a reliable method for calculating heat changes in water and other substances.

References

[1] "Specific Heat Capacity of Water." World Wide School.

[2] "Heat Capacity and Specific Heat." LibreTexts Chemistry.