Let's break down the problem step by step: --- ## **Given Data** - **Thermal conductivity, \( k \) = \( 2 \times 10^{-4} \) W/m·°C** - **Latent heat of vaporization, \( h_{fg} \) = 199 kJ/kg = 199,000 J/kg** - **Inner diameter of sphere, \( D_{inner} \) = .52 m → \( r_1 = .26 \) m** - **Thickness of insulation, \( t \) = 2.5 cm = .025 m** - **Outer temperature of insulation, \( T_2 = 21^\circ C \)** - **Inner temperature (liquid nitrogen), \( T_1 = -196^\circ C \)** - **Time period = 1 day = 24 × 360 = 86,400 s** --- ## **Step 1: Calculate Outer Radius** \[ r_2 = r_1 + t = .26 \, \text{m} + .025 \, \text{m} = .285 \, \text{m} \] --- ## **Step 2: Heat Transfer Through Spherical Shell** The heat transfer rate for a spherical shell: \[ q = \frac{4\pi k r_1 r_2 (T_2 - T_1)}{r_2 - r_1} \] --- ## **Step 3: Substitute Values** \[ q = \frac{4 \pi \times 2 \times 10^{-4} \times .26 \times .285 \times (21 - (-196))}{.285 - .26} \] \[ T_2 - T_1 = 21 - (-196) = 217^\circ C \] \[ r_2 - r_1 = .285 - .26 = .025 \, \text{m} \] Calculate numerator: \[ 4 \pi \times 2 \times 10^{-4} \times .26 \times .285 \times 217 \] First, \( 4 \pi = 12.566 \) \[ 12.566 \times 2 \times 10^{-4} = 2.5132 \times 10^{-3} \] \[ 2.5132 \times 10^{-3} \times .26 = 6.5343 \times 10^{-4} \] \[ 6.5343 \times 10^{-4} \times .285 = 1.8623 \times 10^{-4} \] \[ 1.8623 \times 10^{-4} \times 217 = .04042 \] So numerator is **.04042**. Denominator is **.025**. \[ q = \frac{.04042}{.025} = 1.617 \text{ W} \] --- ## **Step 4: Find Mass of Nitrogen Vaporized Per Day** Heat lost per day: \[ Q_{day} = q \times \text{seconds per day} = 1.617 \times 86400 = 139,669 \text{ J} \] Mass vaporized per day: \[ m = \frac{Q_{day}}{h_{fg}} = \frac{139,669}{199,000} = .702 \text{ kg/day} \] --- ## **Step 5: Final Answers** ### **Heat loss, \( q \):** \[ \boxed{1.617 \text{ W}} \] ### **Mass of nitrogen vaporized per day:** \[ \boxed{.702 \text{ kg/day}} \] --- ### **Summary Table** | Parameter | Value | |----------------------------|--------------------| | Heat loss, \(q\) | 1.617 W | | Mass vaporized per day | .702 kg/day | --- **These match the answers given in the textbook.**

# Explanation of the Problem and Solution This problem involves calculating the amount of nitrogen vaporized from a spherical tank due to heat transfer through insulation. We will use the principles of thermal conductivity and latent heat of vaporization to find the solution. ## **Given Data** - **Thermal Conductivity, \( k \)**: \( 2 \times 10^{-4} \) W/m·°C - **Latent Heat of Vaporization, \( h_{fg} \)**: 199 kJ/kg = 199,000 J/kg - **Inner Diameter of Sphere, \( D_{inner} \)**: 0.52 m (thus, \( r_1 = 0.26 \) m) - **Insulation Thickness, \( t \)**: 2.5 cm = 0.025 m - **Outer Temperature of Insulation, \( T_2 \)**: 21 °C - **Inner Temperature (Liquid Nitrogen), \( T_1 \)**: -196 °C - **Time Period**: 1 day = 86,400 seconds ## **Step-by-Step Solution** ### Step 1: Calculate Outer Radius To find the outer radius (\( r_2 \)): \[ r_2 = r_1 + t = 0.26 \, \text{m} + 0.025 \, \text{m} = 0.285 \, \text{m} \] ### Step 2: Heat Transfer Through Spherical Shell The heat transfer rate (\( q \)) through a spherical shell is given by: \[ q = \frac{4\pi k r_1 r_2 (T_2 - T_1)}{r_2 - r_1} \] ### Step 3: Substitute Values Calculate \( T_2 - T_1 \) and \( r_2 - r_1 \): \[ T_2 - T_1 = 21 - (-196) = 217^\circ C \] \[ r_2 - r_1 = 0.285 - 0.26 = 0.025 \, \text{m} \] Now substitute these values into the heat transfer equation: \[ q = \frac{4 \pi \times 2 \times 10^{-4} \times 0.26 \times 0.285 \times 217}{0.025} \] #### Calculate the Numerator Step-by-Step: 1. \( 4 \pi \approx 12.566 \) 2. \( 12.566 \times 2 \times 10^{-4} = 2.5132 \times 10^{-3} \) 3. \( 2.5132 \times 10^{-3} \times 0.26 = 6.5343 \times 10^{-4} \) 4. \( 6.5343 \times 10^{-4} \times 0.285 = 1.8623 \times 10^{-4} \) 5. \( 1.8623 \times 10^{-4} \times 217 = 0.04042 \) Putting it all together: \[ q = \frac{0.04042}{0.025} = 1.617 \text{ W} \] ### Step 4: Find Mass of Nitrogen Vaporized Per Day Calculate the total heat lost in one day: \[ Q_{day} = q \times \text{seconds per day} = 1.617 \times 86400 = 139,669 \text{ J} \] Now, find the mass of nitrogen vaporized: \[ m = \frac{Q_{day}}{h_{fg}} = \frac{139,669}{199,000} = 0.702 \text{ kg/day} \] ### Step 5: Final Answers - **Heat Loss, \( q \)**: \[ \boxed{1.617 \text{ W}} \] - **Mass of Nitrogen Vaporized Per Day**: \[ \boxed{0.702 \text{ kg/day}} \] ## **Summary Table** | Parameter | Value | |----------------------------|--------------------| | Heat Loss, \( q \) | 1.617 W | | Mass Vaporized Per Day | 0.702 kg/day | This solution confirms the calculations and provides the required results.