During a dry day, a 70.0 kg athlete runs a marathon while the outside temperature is T2 = 33.2 ¢XC. The thermal power produced by the athlete's body is P = 1.05 ¡Ñ 103 W and the skin temperature is T3 = 35.5 ¢XC so that the themal exchange due to convection, radiation and conduction are negligible.
a) What quantity of sweat must be evaporated per second so that the athlete's body temperature will remain constant.
b) The maximum sweat production rate is Ds = 0.220 g s−1. What is the thermal energy increase of the athlete's body per second?
c) How long would it take to increase the body temperature by 1.00 ¢XC if there is no other mechanism of heat loss other than evaporation? The specific heat of the athlete's body is 4.50 ¡Ñ 103 J kg−1 K−1
To find the quantity of sweat that must be evaporated per second, we need to use the equation for heat transfer through evaporation:
Q = m * L,
where Q is the heat transferred, m is the mass of sweat evaporated, and L is the latent heat of evaporation.
a) First, let's find the mass of sweat evaporated per second. We know the thermal power produced by the athlete's body is 1.05 × 10^3 W. The sweat production rate is Ds = 0.220 g/s, which is equivalent to 0.220 × 10^−3 kg/s. We can use the equation for power:
Power = Q / t,
where Q is the heat transferred and t is the time. Rearranging the equation, we have:
Q = Power * t.
From this equation, we can see that heat transferred is equal to the thermal power multiplied by the time.
Next, we need to find the latent heat of evaporation, which is the amount of heat required to evaporate one kilogram of sweat. We can use the specific heat of water (c = 4.18 × 10^3 J/kg·°C) and the latent heat of fusion for water (Lf = 2.26 × 10^6 J/kg). The latent heat of evaporation is given by:
L = Lf + c * (T2 - T3),
where T2 is the outside temperature and T3 is the skin temperature.
Substituting the given values, we can calculate the latent heat of evaporation.
Finally, we can plug in the values for the thermal power, time, and latent heat into the equation for heat transfer:
Q = Power * t = m * L.
Solving for m, we can calculate the mass of sweat evaporated per second.
b) To find the thermal energy increase of the athlete's body per second, we need to use the equation:
Q = mcΔT,
where Q is the heat transferred, m is the mass of the athlete's body, c is the specific heat capacity of the body, and ΔT is the increase in temperature.
The thermal energy increase is given by the product of mass, specific heat, and temperature change.
c) To calculate how long it would take to increase the body temperature by 1.00 ¢XC, we can rearrange the equation used in part b:
t = ΔT / (mc),
where t is the time, ΔT is the increase in temperature, m is the mass of the athlete's body, and c is the specific heat capacity.
By substituting the given values into the equation, we can find the time required to increase the body temperature.
Note: In all calculations, make sure to convert the temperature units to Kelvin (K) by adding 273.15.
To solve this problem, we need to consider the heat transfer from the athlete's body due to evaporation of sweat.
a) The heat transferred by the evaporation of sweat is given by the equation:
Q = m * L
where Q is the heat transferred, m is the mass of the sweat being evaporated, and L is the latent heat of vaporization.
Given that the thermal power produced by the athlete's body is P = 1.05 × 10^3 W, we can equate this to the heat transferred by the evaporation:
P = Q/t
where t is the time taken for the evaporation to occur.
Rearranging the equation gives us:
Q = P * t
Since the body temperature remains constant, the heat transferred by evaporation must equal the heat produced by the body:
P = Q
Substituting Q = P * t, we can solve for t:
t = P/P = 1.05 × 10^3 W / 1.05 × 10^3 W = 1 second
Therefore, the athlete must evaporate 1 gram of sweat per second to maintain a constant body temperature.
b) The thermal energy increase of the athlete's body per second can be calculated by multiplying the mass of sweat being evaporated per second (Ds) by the latent heat of vaporization (L):
Thermal energy increase = Ds * L
Given Ds = 0.220 g/s, we need to convert the mass to kilograms:
Ds = 0.220 g/s = 0.220 × 10^(-3) kg/s
The latent heat of vaporization for water is L = 2.26 × 10^6 J/kg. Substituting these values, we can calculate the thermal energy increase:
Thermal energy increase = 0.220 × 10^(-3) kg/s * 2.26 × 10^6 J/kg = 0.4972 J/s
Therefore, the thermal energy increase of the athlete's body per second is 0.4972 J/s.
c) To calculate the time it would take to increase the body temperature by 1.00 ¢XC using only evaporation as the heat loss mechanism, we need to consider the specific heat capacity of the athlete's body.
The heat energy required to increase the body temperature by 1.00 ¢XC can be calculated using the equation:
Q = m * c * ΔT
where Q is the heat energy, m is the mass of the athlete's body, c is the specific heat capacity, and ΔT is the change in temperature.
Given that the specific heat capacity of the athlete's body is c = 4.50 × 10^3 J/kg K and the mass of the athlete's body is not provided, we will assume it to be 70.0 kg.
Substituting the values:
Q = 70.0 kg * 4.50 × 10^3 J/kg K * 1.00 ¢XC = 315,000 J
The heat energy required for a temperature increase of 1.00 ¢XC is 315,000 J.
Since the heat loss mechanism in this scenario is only evaporation, the heat lost through evaporation must match the heat energy required for the temperature increase:
Q = Thermal energy increase
315,000 J = Ds * L * t
Rearranging the equation to solve for t:
t = (315,000 J) / (Ds * L)
Substituting Ds = 0.220 × 10^(-3) kg/s and L = 2.26 × 10^6 J/kg, we can calculate the time:
t = (315,000 J) / (0.220 × 10^(-3) kg/s * 2.26 × 10^6 J/kg) ≈ 641.5 seconds
Therefore, it would take approximately 641.5 seconds to increase the body temperature by 1.00 ¢XC using only evaporation as the heat loss mechanism.