Question

A \(68-\mathrm{kg}\) man whose average body temperature is \(39^{\circ} \mathrm{C}\) drinks \(1 \mathrm{L}\) of cold water at \(3^{\circ} \mathrm{C}\) in an effort to cool down. Taking the average specific heat of the human body to be \(3.6 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C},\) determine the drop in the average body temperature of this person under the influence of this cold water.

Short answer

Answer: The drop in the average body temperature is approximately 0.613°C.

Step-by-step solution

Identify given values

The problem gives us the following values: - Mass of the man (m): \(68 \ \mathrm{kg}\) - Initial average body temperature (T1): \(39^{\circ} \mathrm{C}\) - Volume of cold water consumed (V): \(1 \ \mathrm{L}\) - Temperature of cold water (T2): \(3^{\circ} \mathrm{C}\) - Average specific heat of the human body (c): \(3.6 \ \mathrm{kJ/kg \cdot^{\circ} C}\)

Convert the volume of water to mass

Assuming that the density of water is approximately \(1 \ \mathrm{kg/L}\), we can calculate the mass of the cold water consumed (m_water) as: \(m_{water} = V \times 1 \ \mathrm{kg/L} = 1 \ \mathrm{kg}\)

Calculate the energy absorbed by the cold water

For the consumed water to reach the body's initial temperature, it will absorb the energy (Q) from the body. We can calculate the energy absorbed by the cold water using the following formula: \(Q = m_{water} \times c_{water} \times (T1 - T2)\) However, we should remember that the specific heat of water is approximately \(4.18 \ \mathrm{kJ/kg \cdot^{\circ} C}\). Now, we can plug in the values: \(Q = 1 \ \mathrm{kg} \times 4.18 \ \mathrm{kJ/kg\cdot^{\circ}C} \times (39^{\circ} \mathrm{C} - 3^{\circ} \mathrm{C}) = 150.36 \ \mathrm{kJ}\)

Calculate the temperature drop for the whole body

We will now calculate the temperature drop (∆T) by equating the energy absorbed by the cold water to the energy released by the body: \(m \times c \times \Delta T = Q\) \(m \times c \times \Delta T = 150.36 \ \mathrm{kJ}\) Then, taking the mass and specific heat of the human body: \(68 \ \mathrm{kg} \times 3.6 \ \mathrm{kJ/kg \cdot^{\circ}C} \times \Delta T = 150.36 \ \mathrm{kJ}\) Now, solving for ∆T: \(\Delta T = \dfrac{150.36 \ \mathrm{kJ}}{68 \ \mathrm{kg} \times 3.6 \ \mathrm{kJ/kg \cdot^{\circ}C}} \approx 0.613^{\circ} \mathrm{C}\) Therefore, the drop in the average body temperature under the influence of the cold water is approximately \(0.613^{\circ} \mathrm{C}\).