Let's break down the problem step by step. --- # **Given Data** - **Mass of coffee, \( m \) = .2 kg** - **Rate of heat transfer, \( \dot{Q} = .1~\text{kW} = 100~\text{W} \)** - **Density, \( \rho = 100~\text{kg/m}^3 \)** - **Specific internal energy, \( u = 4.2~T \) (in kJ/kg, \( T \) in K)** - **Temperature increase, \( \Delta T = 20^\circ C \)** - **Assume coffee is liquid water.** --- ## **(a) System & Assumptions** - **System:** The coffee in the cup. - **Boundary:** The cup boundary (assume no heat loss except microwave energy input). - **Assumptions:** 1. The coffee is well-mixed (uniform temperature). 2. No phase change occurs. 3. No work is done by or on the system, only heat transfer. 4. The specific heat capacity can be approximated as constant over small temperature ranges. 5. Neglect heat losses to the surroundings. --- ## **(b) Rate of Change of Internal Energy** The general energy balance for the system is: \[ \frac{dU}{dt} = \dot{Q} - \dot{W} \] Since there is no work (\( \dot{W} = \)): \[ \frac{dU}{dt} = \dot{Q} \] The internal energy (\( U \)) for a mass is: \[ U = m u \] So, \[ \frac{dU}{dt} = m \frac{du}{dt} \] Thus, \[ m \frac{du}{dt} = \dot{Q} \] --- ## **(c) Time for Temperature Increase** Given \( u = 4.2~T \) (kJ/kg), convert to J/kg: \[ u = 4.2~T \times 100 = 420~T~(\text{J/kg}) \] But usually, specific heat (\( c \)) of water is 4186 J/kg·K, so this is consistent. \[ \frac{du}{dt} = \frac{du}{dT} \cdot \frac{dT}{dt} \] Given \( u = 4.2~T \): \[ \frac{du}{dT} = 4.2~(\text{kJ/kg·K}) = 420~(\text{J/kg·K}) \] So, \[ m \frac{du}{dt} = m \frac{du}{dT} \frac{dT}{dt} = \dot{Q} \] Solving for \( \frac{dT}{dt} \): \[ m \cdot 420 \cdot \frac{dT}{dt} = 100 \] \[ .2 \cdot 420 \cdot \frac{dT}{dt} = 100 \] \[ 840 \cdot \frac{dT}{dt} = 100 \] \[ \frac{dT}{dt} = \frac{100}{840} = .119~\text{K/s} \] To increase temperature by \( \Delta T = 20^\circ C \): \[ \Delta t = \frac{\Delta T}{dT/dt} = \frac{20}{.119} \approx 168~\text{s} \] --- ## **Final Answers** ### **(a) System and Assumptions** - System: The coffee in the cup. - Assumptions: 1. Well-mixed (uniform temperature). 2. No phase change. 3. Only heat transfer, no work. 4. Neglect heat loss. ### **(b) Rate of Change of Internal Energy** \[ \frac{dU}{dt} = \dot{Q} = 100~\text{W} \] ### **(c) Time for 20°C Temperature Increase** \[ \boxed{168~\text{seconds}} \] --- **Summary:** - **Rate of change of internal energy:** \( 100~\text{W} \) - **Time required for a 20°C increase:** \( 168~\text{seconds} \)