To determine how much water is needed to cool a jogger weighing 77.6 kg by 2.78°C, we can use the formula for heat transfer:
Q = mcΔT
Where:
Q = heat energy (in joules)
m = mass (in kg)
c = specific heat capacity (in J/kg°C)
ΔT = change in temperature (in °C)
First, we can calculate the amount of heat energy required to cool the jogger:
Q = 77.6 kg × 3500 J/kg°C × 2.78°C
Calculating this gives us:
Q = 77.6 × 3500 × 2.78 = 75,000 J (approximately)
Next, we need to find out how much water is required to absorb this heat using the latent heat of vaporization of water at body temperature (37°C), which is given as:
Latent heat of vaporization = 2.42 × 10⁶ J/kg
The amount of heat absorbed by the water when it vaporizes can also be calculated using:
Q = mL
Where L is the latent heat of vaporization. To find the mass of the water (m), we rearrange the equation:
m = Q / L
Substituting the values we have:
m = 75,000 J / (2.42 × 10⁶ J/kg)
Calculating this gives us:
m ≈ 0.031 kg or 31 grams
Therefore, to cool the jogger’s body temperature by 2.78°C, approximately 31 grams of water is needed.