Question

The average specific heat of the human body is \(3.6 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C} .\) If the body temperature of an \(80-\mathrm{kg}\) man rises from \(37^{\circ} \mathrm{C}\) to \(39^{\circ} \mathrm{C}\) during strenuous exercise, determine the increase in the thermal energy of the body as a result of this rise in body temperature.

Short answer

Answer: 576000 J

Step-by-step solution

Identify the given values

We are given the following values in the exercise: - Average specific heat of the human body (\(c\)): \(3.6 \mathrm{~kJ/kg \cdot^{\circ} C} \) - Man's mass (\(m\)): \(80 \mathrm{~kg}\) - Initial body temperature (\(T_i\)): \(37^{\circ}\mathrm{C}\) - Final body temperature (\(T_f\)): \(39^{\circ}\mathrm{C}\)

Compute the change in temperature

We need to find \(ΔT\), which represents the change in the body temperature. We can calculate it using the formula: $$ΔT = T_f - T_i$$ Plugging in the values, we get: $$ΔT = 39^{\circ}\mathrm{C} - 37^{\circ}\mathrm{C} = 2^{\circ}\mathrm{C}$$

Convert specific heat to standard units

The given specific heat is in kJ, but we need it in J (joules) for the calculations. So, we'll convert it: $$1 \mathrm{~kJ} = 1000 \mathrm{~J}$$ So, the specific heat in J is: $$3.6 \mathrm{~kJ/kg \cdot^{\circ} C} \times 1000 = 3600 \mathrm{~J/kg \cdot^{\circ} C}$$

Calculate the increase in thermal energy

Now, we can use the formula for heat transfer to find the increase in thermal energy: $$Q = mcΔT$$ Plugging in the values we found earlier, we get: $$Q = (80 \mathrm{~kg})(3600 \mathrm{~J/kg \cdot^{\circ} C})(2^{\circ}\mathrm{C})$$ $$Q = 576000 \mathrm{~J}$$ The increase in the thermal energy of the body as a result of the rise in body temperature is \(576000 \mathrm{~J}\).