91Ó°ÊÓ

The body cools from \(60^{\circ}\mathrm{C}\) to \(50^{\circ}\mathrm{C}\) in 10 minutes. Assuming the room temperature to be \(25^{\circ} \mathrm{C}\). The initial conditions are given as: \(T_0 = 60^{\circ}\mathrm{C}\), \(T_{10} = 50^{\circ}\mathrm{C}\) and \(T_{room} = 25^{\circ}\mathrm{C}\)

Since we are working with the rate of change of temperature with respect to time, we need to convert the given time to hours to make it consistent with the differential equation. So, 10 minutes = \(\frac{10}{60}\) hours = \(\frac{1}{6}\) hours

We need to find the constant \(k\) that governs how fast the body cools. To achieve this, separate the variables in the differential equation as follows: \[ \frac{\mathrm{d}T}{T - T_{room}} = -k\mathrm{d}t \] Now, integrate both sides with the limits given: \[ \int_{T_0}^{T_{10}}\frac{\mathrm{d}T}{T - T_{room}} = -k\int_{0}^{\frac{1}{6}}\mathrm{d}t \] Solving the integrals: \[ \left[\ln |T - T_{room}|\right]_{T_0}^{T_{10}} = \left[-k \cdot t\right]_0^{\frac{1}{6}} \] After substituting the limits: \[ \ln \frac{T_{10} - T_{room}}{T_0 - T_{room}} = -\frac{1}{6}k \] Calculating k: \[ k = 6\cdot\ln\frac{60 - 25}{50 - 25} \] Now we have the cooling constant: \[ k \approx 0.484 \]

We can now use \(k\) and the same method to compute the temperature of the body after the next 10 minutes. Integrate again: \[ \int_{T_{10}}^{T_{20}}\frac{\mathrm{d}T}{T - T_{room}} = -k\int_{\frac{1}{6}}^{\frac{1}{3}}\mathrm{d}t \] Now, solving for \(T_{20}\): \[ \ln \frac{T_{20} - T_{room}}{T_{10} - T_{room}} = -\frac{1}{6}k \] \[ T_{20} - T_{room} = (T_{10} - T_{room})e^{-\frac{1}{6}k} \] \[ T_{20} = (T_{10} - T_{room})e^{-\frac{1}{6}k} + T_{room} \] Plug in the values: \[ T_{20} = (50-25)e^{-\frac{1}{6}(0.484)} + 25 \] Calculating: \[ T_{20} \approx 40^{\circ}\mathrm{C} \] So the temperature of the body at the end of the next 10 minutes will be around \(40^{\circ}\mathrm{C}\). The correct answer is (B) \(40^{\circ} \mathrm{C}\).