Certainly! Here is the step-by-step solution: --- ###Step 1 Identify the Relevant (Newton’s Law of Cooling)** **Explanation:** The problem involves cooling, with a hot cup of coffee cooling in a room of constant temperature. The temperature change over time follows **Newton's Law of Cooling**, which states: \[ \frac{dT}{dt} = -k(T - T_{\text{env}}) \] where: - \( T \) = temperature of the object (coffee) - \( T_{\text{env}} \) = ambient (room) temperature (20°C) - \( k \) = cooling constant - \( t \) = time --- ### **Step 2: Set Up Equation and Solve for Time** We are given: - Initial coffee temperature, \( T_ = 70^\circ C \) - Room temperature, \( T_{\text{env}} = 20^\circ C \) - Final temperature (drinking temp), \( T = 50^\circ C \) The solution to Newton’s Law is: \[ T(t) = T_{\text{env}} + (T_ - T_{\text{env}})e^{-kt} \] Plug in values: \[ 50 = 20 + (70-20)e^{-kt} \] \[ 50 = 20 + 50e^{-kt} \] \[ 30 = 50e^{-kt} \] \[ \frac{30}{50} = e^{-kt} \] \[ .6 = e^{-kt} \] \[ \ln(.6) = -kt \] \[ t = \frac{-\ln(.6)}{k} \] **Explanation:** This equation gives the time required for the coffee to cool from 70°C to 50°C. The time depends on the cooling constant \( k \), which must be determined experimentally or given. If \( k \) is not given, the answer should be left in terms of \( k \). --- ### **Final Answer** \[ \boxed{t = \frac{-\ln(.6)}{k}} \] **This is the time required for the coffee to cool from 70°C to 50°C in a room at 20°C, according to Newton’s Law of Cooling.**