SP3. The initial temperature of 140 g of ice is \( -22^{\circ} \mathrm{C} \). The specific heat capacity of ice is \( 0.5 \mathrm{cal} / \mathrm{g} \cdot \mathrm{C}^{\circ} \), and water's is \( 1 \mathrm{cal} / \mathrm{g} \cdot \mathrm{C}^{\circ} \). The latent heat of fusion of water is \( 80 \mathrm{cal} / \mathrm{g} \). a. How much heat is required to raise the ice to \( 0^{\circ} \mathrm{C} \) ? b. How much additional heat is required to completely melt the ice after it has reached \( 0^{\circ} \mathrm{C} \) ? c. How much additional heat is required to heat the water (obtained by melting the ice) to \( 27^{\circ} \mathrm{C} \) ? d. What is the total heat that must be added to convert the 140 g of ice at \( -22^{\circ} \mathrm{C} \) to water at \( +27^{\circ} \mathrm{C} \) ? e. Examine your answers to parts a, b, and c . Which takes the most energy: heating ice to the melting point, melt- ing the ice, or heating the cold water to the final temperature? f. Can we find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 140 g of water by \( 49^{\circ} \mathrm{C} \) ? Explain.

To solve this problem, we will break it down into parts as requested. We will use the following formulas: 1. **Heat required to change temperature**: \[ Q = m \cdot c \cdot \Delta T \] where \( Q \) is the heat, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. 2. **Heat required for phase change (melting)**: \[ Q = m \cdot L \] where \( L \) is the latent heat of fusion. ### a. Heat required to raise the ice to \( 0^{\circ} \mathrm{C} \) Given: - Mass of ice, \( m = 140 \, \text{g} \) - Specific heat capacity of ice, \( c = 0.5 \, \text{cal/g} \cdot \text{C}^{\circ} \) - Initial temperature, \( T_i = -22^{\circ} \mathrm{C} \) - Final temperature, \( T_f = 0^{\circ} \mathrm{C} \) The change in temperature, \( \Delta T = T_f - T_i = 0 - (-22) = 22^{\circ} \mathrm{C} \). Now, we can calculate the heat required: \[ Q_a = m \cdot c \cdot \Delta T = 140 \cdot 0.5 \cdot 22 \] ### b. Heat required to completely melt the ice Given: - Latent heat of fusion, \( L = 80 \, \text{cal/g} \) Now, we can calculate the heat required to melt the ice: \[ Q_b = m \cdot L = 140 \cdot 80 \] ### c. Heat required to heat the water to \( 27^{\circ} \mathrm{C} \) Given: - Specific heat capacity of water, \( c = 1 \, \text{cal/g} \cdot \text{C}^{\circ} \) - Initial temperature of water, \( T_i = 0^{\circ} \mathrm{C} \) - Final temperature, \( T_f = 27^{\circ} \mathrm{C} \) The change in temperature, \( \Delta T = T_f - T_i = 27 - 0 = 27^{\circ} \mathrm{C} \). Now, we can calculate the heat required: \[ Q_c = m \cdot c \cdot \Delta T = 140 \cdot 1 \cdot 27 \] ### d. Total heat required The total heat \( Q_{total} \) is the sum of the heats calculated in parts a, b, and c: \[ Q_{total} = Q_a + Q_b + Q_c \] ### e. Energy comparison We will compare the values of \( Q_a \), \( Q_b \), and \( Q_c \) to determine which takes the most energy. ### f. Total heat calculation explanation We will discuss whether we can find the total heat simply by computing the heat required to melt the ice and adding the heat required to raise the temperature of the resulting water. Now, let's perform the calculations for parts a, b, and c. Calculate the value by following steps: - step0: Calculate: \(140\times 80\) - step1: Multiply the numbers: \(11200\) Calculate or simplify the expression \( 140 * 0.5 * 22 \). Calculate the value by following steps: - step0: Calculate: \(140\times 0.5\times 22\) - step1: Multiply the terms: \(70\times 22\) - step2: Multiply the numbers: \(1540\) Calculate or simplify the expression \( 140 * 1 * 27 \). Calculate the value by following steps: - step0: Calculate: \(140\times 1\times 27\) - step1: Rewrite the expression: \(140\times 27\) - step2: Multiply the numbers: \(3780\) ### Calculated Results 1. **Heat required to raise the ice to \( 0^{\circ} \mathrm{C} \)**: \[ Q_a = 1540 \, \text{cal} \] 2. **Heat required to completely melt the ice**: \[ Q_b = 11200 \, \text{cal} \] 3. **Heat required to heat the water to \( 27^{\circ} \mathrm{C} \)**: \[ Q_c = 3780 \, \text{cal} \] ### d. Total heat required Now, we can calculate the total heat required: \[ Q_{total} = Q_a + Q_b + Q_c = 1540 + 11200 + 3780 \] Let's calculate \( Q_{total} \). Calculate the value by following steps: - step0: Calculate: \(1540+11200+3780\) - step1: Add the numbers: \(16520\) ### Total Heat Required The total heat required to convert the 140 g of ice at \( -22^{\circ} \mathrm{C} \) to water at \( +27^{\circ} \mathrm{C} \) is: \[ Q_{total} = 16520 \, \text{cal} \] ### e. Energy Comparison Now, let's compare the energies calculated in parts a, b, and c: - Heating ice to the melting point: \( Q_a = 1540 \, \text{cal} \) - Melting the ice: \( Q_b = 11200 \, \text{cal} \) - Heating the cold water to the final temperature: \( Q_c = 3780 \, \text{cal} \) From these values, we can see that: - The most energy is required for **melting the ice** (\( Q_b = 11200 \, \text{cal} \)), which is significantly higher than the other two processes. ### f. Total Heat Calculation Explanation To find the total heat, we cannot simply compute the heat required to melt the ice and add the amount of heat required to raise the temperature of 140 g of water by \( 49^{\circ} \mathrm{C} \) (from \( 0^{\circ} \mathrm{C} \) to \( 27^{\circ} \mathrm{C} \)). This is because the heat required to raise the temperature of the water includes the energy needed to first melt the ice. The total heat calculation must account for: 1. Heating the ice to \( 0^{\circ} \mathrm{C} \) 2. Melting the ice at \( 0^{\circ} \mathrm{C} \) 3. Heating the resulting water from \( 0^{\circ} \mathrm{C} \) to \( 27^{\circ} \mathrm{C} \) Thus, all three steps are necessary to accurately determine the total heat required.