Question

Calculate the amount of heat energy transferred when 5.0 kg of water is cooled from 80°C to 20°C, assuming a specific level of specific heat capacity of 4.18 J/g°C.

Answer 1

To calculate the amount of heat energy transferred during the cooling of water, we can use the formula:

\[
Q = mc\Delta T
\]

where:
- \( Q \) = heat energy transferred (in joules)
- \( m \) = mass of the substance (in grams)
- \( c \) = specific heat capacity (in J/g°C)
- \( \Delta T \) = change in temperature (in °C)

Given:
- Mass of water: \( m = 5.0 \, \text{kg} = 5000 \, \text{g} \) (since 1 kg = 1000 g)
- Specific heat capacity: \( c = 4.18 \, \text{J/g°C} \)
- Initial temperature: \( T_i = 80°C \)
- Final temperature: \( T_f = 20°C \)
- Change in temperature: \( \Delta T = T_f - T_i = 20°C - 80°C = -60°C \)

Now substituting the values into the formula:

\[
Q = (5000 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (-60 \, \text{°C})
\]

Calculating:

\[
Q = 5000 \times 4.18 \times -60
\]

\[
Q = 5000 \times 4.18 \times -60 = 5000 \times -250.8 = -1254000 \, \text{J}
\]

Since heat energy transferred out of the system is typically reported as a positive value, we can say:

\[
Q = 1254000 \, \text{J}
\]

Thus, the amount of heat energy transferred when 5.0 kg of water is cooled from 80°C to 20°C is **1,254,000 J** or **1.25 MJ**.