Tony, a 110 kg football player has a fever of 39 degrees C. What mass of water must evaporate from his body to cool him to 37 degrees C?
To determine the mass of water that must evaporate from Tony's body to cool him to 37 degrees C, we need to use the concept of specific heat and the latent heat of vaporization. Here's how we can calculate it step by step:
Step 1: Calculate the heat lost by Tony's body when cooling from 39 degrees C to 37 degrees C.
To do this, we'll use the following formula: Q = mcΔT, where Q is the heat lost, m is the mass, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C (joules per gram-degree Celsius).
So, the heat lost (Q) can be calculated as: Q = mcΔT = (110 kg) × (4.18 J/g°C) × (39 - 37) °C.
Step 2: Calculate the heat required to evaporate the water from Tony's body.
The heat required to evaporate water is known as the latent heat of vaporization, which is approximately 2260 J/g (joules per gram).
To calculate the heat required (Q') to evaporate the water, we'll use the following formula: Q' = mL, where L is the latent heat of vaporization.
Step 3: Equate Q and Q' to find the mass of water (m) that must evaporate from Tony's body.
Since Q and Q' represent the same amount of heat, we can write the equation: mcΔT = mL.
Now we can solve for the mass of water (m) by rearranging the equation: m = ( mL ) / ( cΔT ).
Let's plug in the values into the equation and calculate the mass of water required to evaporate:
m = ( (2260 J/g) / (4.18 J/g°C) ) × ( (37 - 39) °C )
m = 542.58 g
Therefore, approximately 542.58 grams of water must evaporate from Tony's body to cool him from 39 degrees C to 37 degrees C.