Let's solve the problem **step-by-step** as requested. ## Step 1: Understanding the System - **Thermometer Response**: First-order differential equation with time constant τ = .2 min. - **Initial Temperature**: Both water bath and thermometer are at 20°C. - **At t = **: Water bath temperature is suddenly increased (by mixing with hot water) and remains constant for t > . ## Step 2: Temperature of Water Bath After Mixing (Part a) At t > , the new temperature of the water bath is found by mixing equal volumes of water at 20°C and 100°C: \[ T_{\text{bath}} = \frac{1 \text{ L} \times 20^\circ\text{C} + 1 \text{ L} \times 100^\circ\text{C}}{2 \text{ L}} = \frac{20 + 100}{2} = 60^\circ\text{C} \] **Answer (a):** The temperature of the water bath at time \( t > \) is **60°C**. --- ## Step 3: Thermometer Response Equation The first-order response of the thermometer is: \[ \tau \frac{dT_m}{dt} + T_m = T \] where: - \( \tau = .2 \) min - \( T_m \) = thermometer reading - \( T \) = water bath temperature (step change from 20°C to 60°C at t = ) Initially, \( T_m() = 20^\circ\text{C} \). The solution for a step change is: \[ T_m(t) = T_{\text{final}} + (T_{\text{initial}} - T_{\text{final}}) e^{-t/\tau} \] Substitute the values: \[ T_m(t) = 60 + (20 - 60) e^{-t / .2} \] \[ T_m(t) = 60 - 40 e^{-5t} \] --- ## Step 4: Find Differences at Specific Times **For (b):** At \( t = .1 \) min \[ T_m(.1) = 60 - 40 e^{-5 \times .1} = 60 - 40 e^{-.5} \] \[ e^{-.5} \approx .6065 \] \[ T_m(.1) = 60 - 40 \times .6065 = 60 - 24.26 = 35.74^\circ\text{C} \] Difference: \[ \Delta T = T - T_m(.1) = 60 - 35.74 = 24.26^\circ\text{C} \] --- **For (c):** At \( t = 1. \) min \[ T_m(1.) = 60 - 40 e^{-5 \times 1.} = 60 - 40 e^{-5} \] \[ e^{-5} \approx .00674 \] \[ T_m(1.) = 60 - 40 \times .00674 = 60 - .2696 = 59.73^\circ\text{C} \] Difference: \[ \Delta T = 60 - 59.73 = .27^\circ\text{C} \] --- ## Step 5: Plot \( T_m \) as a Function of Time (Part d) The function is: \[ T_m(t) = 60 - 40 e^{-5t} \] This is an exponential rise from 20°C at \( t = \) to 60°C as \( t \to \infty \). You can sketch this by plotting \( T_m(t) \) for \( t \) from to 1. min. - At \( t = \): \( T_m = 20^\circ\text{C} \) - At \( t = .1 \): \( T_m \approx 35.74^\circ\text{C} \) - At \( t = .2 \): \( T_m \approx 43.87^\circ\text{C} \) - At \( t = .5 \): \( T_m \approx 56.79^\circ\text{C} \) - At \( t = 1. \): \( T_m \approx 59.73^\circ\text{C} \) The curve approaches 60°C asymptotically. --- ## Final Answers (Summary) (a) The temperature of the water bath at time \( t > \) is **60°C**. (b) The difference between the measured temperature and the bath temperature at \( t = .1 \) min is **24.26°C**. (c) The difference at \( t = 1. \) min is **.27°C**. (d) The measured temperature as a function of time is \( T_m(t) = 60 - 40 e^{-5t} \), which rises exponentially from 20°C to 60°C.