While running, a 70.0 kg student generates thermal energy at a rate of 1200 W. To maintain a constant body temperature of 37.0 C, this energy must be removed by perspiration or other mechanisms.
If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Protein structures in the body are damaged irreversibly if the body temperature rises to 44.0 C or above. The specific heat capacity of a typical human body is 3480 J/(kg*K) slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heat capacities.)
t=?? seconds
I believe this is a Quantity of Heat problem, and the equation that I found that I think would work is:
(Delta)Temp/(Delta)time=
(Q/(delta)t)/mc
(44-37)/t=1200/t /(70*3480)
The answer I got using this was incorrect though...
You are correct that this is a quantity of heat problem. However, the mistake in your calculation is the interpretation of the heat flow rate.
The equation you are using, (Delta)Temp/(Delta)time = Q/(delta)t)/mc, is the correct equation to use. However, the value of Q in this equation should be the rate at which thermal energy is generated, not the total amount of heat produced.
Given that the student generates thermal energy at a rate of 1200 W, the correct value for Q in the equation is 1200 W. So the equation becomes:
(44 - 37) / t = 1200 / (70 * 3480)
Simplifying this equation, we get:
7 / t = 0.004
Solving for t, we find:
t = 7 / 0.004
t ≈ 1750 seconds
Therefore, a student could run for approximately 1750 seconds (or about 29 minutes and 10 seconds) before irreversible body damage occurs if the mechanisms for heat dissipation fail.
To solve this problem, we can use the formula for the rate of heat transfer:
Q = mcΔT
Where:
Q is the heat transferred (in Joules)
m is the mass of the student (in kg)
c is the specific heat capacity of the body (in J/(kg*K))
ΔT is the change in temperature (in K)
In this case, we know the rate of thermal energy produced by the student (1200 W) and the specific heat capacity of the body (3480 J/(kg*K)). However, we need to find the amount of time the student can run before irreversible body damage occurs.
To find the time, we need to consider that the thermal energy generated must be equal to the heat transferred:
Q = 1200 W * t (where t is the time in seconds)
We also need to find the change in temperature (ΔT) that would occur if the heat could not flow out of the student's body. To do this, we can use the formula:
ΔT = Q / (m * c)
Substituting the known values, we have:
ΔT = (1200 W * t) / (70 kg * 3480 J/(kg*K))
Now, we need to set up an equation to find the time (t) at which the change in temperature (ΔT) reaches or exceeds 44 K:
(44 K - 37 K) = (1200 W * t) / (70 kg * 3480 J/(kg*K))
Simplifying the equation, we have:
7 K = (1200 W * t) / (70 kg * 3480 J/(kg*K))
To solve for t, we can rearrange the equation:
t = (7 K * 70 kg * 3480 J/(kg*K)) / 1200 W
Evaluating the expression, we find:
t ≈ 17,900 seconds
Therefore, a student can run for approximately 17,900 seconds, or about 4 hours and 58 minutes, before irreversible body damage occurs if the mechanisms to remove heat fail.
I can't tell from your equation what you did; however,
q = mass x specific heat x delta T.
q = 70 kg x 3480 J/kg*K x 7 = ??
Then q x (1 sec/1200 J) = time in sec.
Check my thinking. 1421 seconds (about 24 minutes)