Question

Estimate the mass of water that must evaporate from the skin to cool the body by 1.50 °C . Assume a body mass of

87.0 kg and assume that the specific heat capacity of the body is 4.00 J/g⋅ °C

Answer 1

To estimate the mass of water that must evaporate from the skin to cool the body by 1.50 °C, we can use the formula Q = mL, where Q is the heat required, m is the mass of water, and L is the latent heat of vaporization.

Step-by-step explanation:

To estimate the mass of water that must evaporate from the skin to cool the body by 1.50 °C, we can use the formula Q = mL, where Q is the heat required, m is the mass of water, and L is the latent heat of vaporization. Since we are given the body mass as 87.0 kg, we can convert it to grams by multiplying by 1000. Then, we can calculate the heat required using the formula Q = mcΔT, where c is the specific heat capacity. Solving for m gives us the mass of water that must evaporate.

Using the given specific heat capacity of 4.00 J/g⋅°C, the heat required to cool the body by 1.50 °C can be calculated as follows:

1. Convert the body mass to grams: 87.0 kg * 1000 g/kg = 87,000 g
2. Calculate the heat required: Q = mcΔT = (87,000 g)(4.00 J/g⋅°C)(1.50 °C) = 522,000 J

Therefore, the mass of water that must evaporate from the skin to cool the body by 1.50 °C is 522,000 grams.