Compare the heat energy in a teaspoon of boiling water and a swimming pool full of room temperature water.

1 answer

To compare the heat energy in a teaspoon of boiling water and a swimming pool full of room temperature water, we need to consider both the specific heat capacity of water and the masses involved.

  1. Teaspoon of Boiling Water:

    • A teaspoon holds about 4.9 milliliters (mL) or approximately 4.9 grams (since the density of water is about 1 g/mL).
    • Boiling water is at 100°C.
  2. Swimming Pool of Room Temperature Water:

    • A standard swimming pool is typically around 50,000 liters or more, which is about 50,000,000 grams.
    • Room temperature water is usually around 20°C.

Calculating Heat Energy

The heat energy (Q) in a mass of water can be calculated using the formula:

\[ Q = mc\Delta T \]

Where:

  • \( m \) = mass of the water (in grams)
  • \( c \) = specific heat capacity of water (approximately \( 4.18 , \text{J/g°C} \))
  • \( \Delta T \) = change in temperature (in °C)

1. Heat Energy in the Teaspoon of Boiling Water

Assuming the boiling water is cooling down to room temperature (from 100°C to 20°C):

  • Mass \( m = 4.9 , \text{g} \)
  • \( \Delta T = 100 - 20 = 80 , \text{°C} \)

Calculating \( Q \):

\[ Q_{teaspoon} = 4.9 \times 4.18 \times 80 \] \[ Q_{teaspoon} ≈ 1,636.64 , \text{J} \]

2. Heat Energy in the Swimming Pool

Assuming the pool is at room temperature:

  • Mass \( m = 50,000,000 , \text{g} \)
  • \( \Delta T = 20 - 20 = 0 , \text{°C} \) (no change in temperature)

Calculating \( Q \):

Since the temperature doesn't change, the heat energy to maintain this temperature doesn’t change, but we can calculate the total energy:

As the water is just at room temperature, we focus on the energy present, which, while not cooling down, can be thought of as the energy stored at that temperature.

\[ Q_{pool} = 50,000,000 \times 4.18 \times 0 = 0 , \text{J} \]

However, to compare the absolute energy stored at 20°C, we can just note that it's already at equilibrium, and thus doesn't change.

Comparison

  • The boiling water has approximately 1,636.64 J of heat energy when cooling to room temperature.
  • The swimming pool, while not undergoing a temperature change, contains a vastly greater volume of water, and any interaction with a heating source would involve calculations of the total thermal energy present.

Thus, the significant factor is:

  • While a teaspoon of boiling water contains the heat energy of about 1,636.64 J, the swimming pool, albeit at equal room temperature, contains a tremendous amount of thermal energy due to its mass (around 50,000,000 grams).

So, the pool has much more total heat energy, even if it's at a cooler temperature overall. In simpler terms, the total energy content of the swimming pool is far greater than that of a single teaspoon of boiling water despite the differences in temperature.