Let's break the problem into two steps: ### Step 1: Heating the Water from 14°C to 100°C The heat required to change the temperature is given by: \[ q_1 = m \cdot c \cdot \Delta T \] where: - \( m = 900.0 \) g (mass of water) - \( c = 1.00 \) cal/g°C (specific heat of water) - \( \Delta T = 100.0^\circ\text{C} - 14.0^\circ\text{C} = 86.0^\circ\text{C} \) So, \[ q_1 = 900.0 \times 1.00 \times 86.0 = 77,400 \text{ cal} \] --- ### Step 2: Converting Water at 100°C to Steam at 100°C The heat required to vaporize water: \[ q_2 = m \cdot L_v \] where: - \( L_v = 540 \) cal/g (heat of vaporization) So, \[ q_2 = 900.0 \times 540 = 486,000 \text{ cal} \] --- ### Total Heat Required \[ Q_{\text{total}} = q_1 + q_2 = 77,400 + 486,000 = \boxed{563,400 \text{ cal}} \] --- ### **Final Answer** \[ \boxed{563,400 \text{ cal}} \] This is the total heat required to change 900.0 g of water at 14°C to steam at 100°C.

Let's carefully analyze and solve the problem step-by-step, providing detailed calculations and explanations. --- ### **Problem Overview:** - **Mass of water, \(m\):** 900.0 g - **Initial temperature of water, \(T_i\):** 14.0°C - **Final temperature of water before vaporization, \(T_f\):** 100.0°C - **Heat of vaporization of water, \(L_v\):** 540 cal/g --- ### **Objective:** Calculate the total heat energy required to: 1. **Heat the water from 14°C to 100°C**. 2. **Convert the water at 100°C into steam at 100°C**. --- ### **Step 1: Heating the water from 14°C to 100°C** The heat energy required to raise the temperature of a substance is given by: \[ q_1 = m \times c \times \Delta T \] Where: - \(m = 900.0\, \text{g}\) - \(c = 1.00\, \text{cal/g°C}\) (specific heat capacity of water) - \(\Delta T = T_f - T_i = 100.0°C - 14.0°C = 86.0°C\) **Calculations:** \[ q_1 = 900.0\, \text{g} \times 1.00\, \text{cal/g°C} \times 86.0°C \] \[ q_1 = 900.0 \times 86.0 = 77,400\, \text{cal} \] **Explanation:** This is the energy needed to raise the temperature of 900 g of water from 14°C to 100°C. --- ### **Step 2: Vaporizing the water at 100°C** The energy required to convert water at 100°C into steam at 100°C is: \[ q_2 = m \times L_v \] Where: - \(L_v = 540\, \text{cal/g}\) **Calculations:** \[ q_2 = 900.0\, \text{g} \times 540\, \text{cal/g} = 486,000\, \text{cal} \] **Explanation:** This is the latent heat energy needed to change 900 g of water at boiling point into steam. --- ### **Step 3: Total heat energy required** Total heat energy: \[ Q_{\text{total}} = q_1 + q_2 \] \[ Q_{\text{total}} = 77,400\, \text{cal} + 486,000\, \text{cal} = 563,400\, \text{cal} \] --- ### **Final Answer:** \[ \boxed{ \textbf{Total heat required} = \mathbf{563,400\, \text{cal}} } \] **Summary:** - Heating water from 14°C to 100°C consumes **77,400 cal**. - Vaporizing water at 100°C consumes **486,000 cal**. - **Total energy needed = 563,400 cal**. This is the amount of heat energy required to convert 900 g of water at 14°C into steam at 100°C.