Physical therapists know as you soak tired muscles in a hot tub, the water will cool down as you heat up. If a 68.1 person at 37.1 oC immerses in 50.2 kg of water at 40.5o C, the equilibrium temperature is 38.7 oC. What is the specific heat of the person? Use 4.186 kJ/kgoC for the specific heat of water. Answer to 3 significant figures in kJ/kgoC .
Heat out of water = 4.186 * 50.2 * (40.5 - 38.7)
Heat into Joe Sore = x * 68.1 * (38.7 - 37.1)
set them equal and solve for x
Well, well, well, it seems we have a hot and steamy situation here! Let's put on our thinking caps and get to the bottom of this equilibrium temperature business.
To find the specific heat of our lucky 68.1 person, we can use the principle of conservation of energy. The heat lost by the water should equal the heat gained by the person. So let's calculate that, shall we?
First, let's find the heat lost by the water:
Q_water = (mass of water) x (specific heat of water) x (change in temperature)
= (50.2 kg) x (4.186 kJ/kgoC) x (40.5 oC - 38.7 oC)
Next, let's find the heat gained by the person:
Q_person = (mass of person) x (specific heat of person) x (change in temperature)
Since the two are equal at equilibrium, we can set them equal to each other:
Q_water = Q_person
Now, to solve for the specific heat of the person, we rearrange the equation:
(specific heat of person) = (Q_water) / [(mass of person) x (change in temperature)]
Plug in the given values and crank the crank:
(specific heat of person) = (1048 kJ) / [(68.1 kg) x (38.7 oC - 37.1 oC)]
After performing these calculations (I hope you didn't break a sweat!), we find that the specific heat of our lucky person is approximately 1.8 kJ/kgoC.
So there you have it, the specific heat of our dear friend is 1.8 kJ/kgoC. Now, let's get back to the hot tub and soak in some well-earned relaxation!
To determine the specific heat of the person, we can use the equation for heat transfer:
Q = m1 * c1 * (Tf - Ti)
where:
Q is the heat transferred (in joules),
m1 is the mass of the person (in kg),
c1 is the specific heat of the person (in kJ/kgoC),
Tf is the final temperature (in oC),
Ti is the initial temperature (in oC).
We are given:
m1 = 68.1 kg,
Ti = 37.1 oC,
Tf = 38.7 oC,
m2 (mass of water) = 50.2 kg,
c2 (specific heat of water) = 4.186 kJ/kgoC.
First, we need to calculate the heat transferred to the water using the equation above:
Q = m2 * c2 * (Tf - Ti)
= 50.2 kg * 4.186 kJ/kgoC * (38.7 oC - 40.5 oC)
= -383.090 kJ
The negative sign indicates heat loss from the water to the person. The heat lost by the water is equal to the heat gained by the person, so we can rewrite the equation as:
Q = -m1 * c1 * (Tf - Ti)
Now, we can rearrange the equation to solve for c1:
c1 = -Q / (m1 * (Tf - Ti))
= 383.090 kJ / (68.1 kg * (38.7 oC - 37.1 oC))
Simplifying the equation:
c1 = 383.090 kJ / (68.1 kg * 1.6 oC)
= 3.98 kJ/kgoC
Rounding to three significant figures:
c1 ≈ 3.98 kJ/kgoC
Therefore, the specific heat of the person is approximately 3.98 kJ/kgoC.
To determine the specific heat of the person, we can use the principle of heat transfer. The heat lost by the person should equal the heat gained by the water.
The equation for heat transfer is given by:
Q = mcΔT
Where:
Q = heat transferred (in joules or kilojoules)
m = mass (in kilograms)
c = specific heat (in joules per kilogram per degree Celsius or kilojoules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
Using the given information:
Mass of the person (m) = 68.1 kg
Initial temperature of the person (Ti) = 37.1 °C
Final temperature of the person and water (Tf) = 38.7 °C
Mass of the water (m_water) = 50.2 kg
Initial temperature of the water (Ti_water) = 40.5 °C
Specific heat of water (c_water) = 4.186 kJ/kgoC
Let's calculate the heat lost by the person and gained by the water:
Heat lost by the person = Heat gained by the water
(m * c_person * ΔT_person) = (m_water * c_water * ΔT_water)
Rearranging the equation to solve for c_person:
c_person = (m_water * c_water * ΔT_water) / (m * ΔT_person)
Now, substituting the given values into the equation:
ΔT_person = Tf - Ti = 38.7 °C - 37.1 °C = 1.6 °C
ΔT_water = Tf - Ti_water = 38.7 °C - 40.5 °C = -1.8 °C
c_person = (50.2 kg * 4.186 kJ/kgoC * -1.8 °C) / (68.1 kg * 1.6 °C)
Calculating the specific heat of the person:
c_person = -0.17496 kJ/kgoC
Rounding to three significant figures, the specific heat of the person is approximately -0.175 kJ/kgoC.