Let's break down the problem step by step: ### Given Data: - **Mass of water, \( m \):** \( 120\,\text{g} = 0.120\,\text{kg} \) - **Initial temperature, \( T_1 \):** \( 34.0^\circ\text{C} \) - **Final temperature, \( T_2 \):** \( 73.0^\circ\text{C} \) - **Power supplied, \( P \):** \( 56.0\,\text{W} \) - **Specific heat capacity of water, \( c \):** \( 4.186 \times 10^3 \,\text{J/(kg}\cdot\text{°C)} \) - **Assume perfect efficiency** (all power goes to heating the water). --- ### Step 1: Calculate the heat required (\( Q \)) \[ Q = mc\Delta T \] where \[ \Delta T = T_2 - T_1 = 73.0 - 34.0 = 39.0^\circ\text{C} \] Plug in the values: \[ Q = (0.120\,\text{kg}) \times (4.186 \times 10^3\,\text{J/kg}^\circ\text{C}) \times (39.0^\circ\text{C}) \] Calculate: - \( 4.186 \times 10^3 = 4186 \) - \( 0.120 \times 4186 = 502.32 \) - \( 502.32 \times 39.0 = 19,590.48 \) So, \[ Q = 19,590.48\,\text{J} \] --- ### Step 2: Calculate the time required (\( t \)) \[ t = \frac{Q}{P} \] \[ t = \frac{19,590.48\,\text{J}}{56.0\,\text{W}} \] \[ t = 349.83\,\text{s} \] --- ### **Final Answer (rounded appropriately):** \[ \boxed{350\,\text{seconds}} \] --- **It will take approximately 350 seconds to raise the temperature of 120 g of water from 34.0°C to 73.0°C using 56.0 W of power.** --- If you want to try with the other set of numbers (\(250\,\text{g}\), \(80\,\text{W}\), \(12^\circ\text{C}\) to \(75^\circ\text{C}\)), let me know!

Let's analyze the second set of data step by step: ### Given Data: - **Mass of water, \( m \):** 250 g = 0.250 kg - **Power supplied, \( P \):** 80 W - **Initial temperature, \( T_1 \):** 12°C - **Final temperature, \( T_2 \):** 75°C - **Specific heat capacity of water, \( c \):** 4.186 × 10³ J/(kg·°C) ### Step 1: Calculate the temperature change (\( \Delta T \)) \[ \Delta T = T_2 - T_1 = 75 - 12 = 63^\circ\text{C} \] ### Step 2: Calculate the heat required (\( Q \)) \[ Q = mc\Delta T \] Substituting the values: \[ Q = 0.250\,\text{kg} \times 4.186 \times 10^3\,\text{J/(kg·°C)} \times 63^\circ\text{C} \] Calculate: - \( 4.186 \times 10^3 = 4186 \) - \( 0.250 \times 4186 = 1046.5 \) - \( 1046.5 \times 63 = 65,919.5\,\text{J} \) ### Step 3: Calculate the time (\( t \)) \[ t = \frac{Q}{P} = \frac{65,919.5\,\text{J}}{80\,\text{W}} \approx 824.0\,\text{s} \] ### **Final answer:** It takes approximately **824 seconds** to heat 250 g of water from 12°C to 75°C with 80 W power. --- **Summary:** | Parameter | Value | |------------|--------| | Time needed | **824 seconds** (~13.7 minutes) | Let me know if you'd like any further clarifications!

Certainly! Here's a detailed, step-by-step explanation for both questions, including the reasoning behind each step: --- ## First Question: **Given Data:** - Mass of water, \( m = 120\,g = 0.120\,kg \) - Initial temperature, \( T_1 = 34.0^\circ C \) - Final temperature, \( T_2 = 73.0^\circ C \) - Power supplied, \( P = 56.0\,W \) - Specific heat capacity of water, \( c = 4.186 \times 10^3\,J/(kg \cdot ^\circ C) \) --- ### Step 1: Find the temperature change (\( \Delta T \)) \[ \Delta T = T_2 - T_1 = 73.0^\circ C - 34.0^\circ C = 39.0^\circ C \] **Reasoning:** The amount of heat needed depends on how much we want to raise the temperature, which is the difference between initial and final temperatures. --- ### Step 2: Calculate the heat energy required (\( Q \)) \[ Q = mc\Delta T \] **Why this formula?** The heat energy needed to raise the temperature of a substance is proportional to its mass, its specific heat capacity, and the temperature change. Plugging in the numbers: \[ Q = 0.120\,kg \times 4186\,J/(kg \cdot ^\circ C) \times 39.0^\circ C \] Calculations: - \( 0.120 \times 4186 = 502.32 \) - \( 502.32 \times 39.0 \approx 19,590.48\,J \) **Result:** \[ Q \approx 19,590.48\,J \] --- ### Step 3: Calculate the time required (\( t \)) \[ t = \frac{Q}{P} \] **Why?** Power is the rate of energy transfer; dividing total energy by power gives the time. \[ t = \frac{19,590.48\,J}{56.0\,W} \approx 349.83\,s \] **Rounded to:** \[ \boxed{350\,seconds} \] **Answer:** It takes approximately **350 seconds** to heat the water from 34.0°C to 73.0°C with 56.0 W. --- ## Second Question: **Given Data:** - Mass of water, \( m = 250\,g = 0.250\,kg \) - Power, \( P = 80\,W \) - Initial temperature, \( T_1 = 12^\circ C \) - Final temperature, \( T_2 = 75^\circ C \) - Specific heat capacity, \( c = 4.186 \times 10^3\,J/(kg \cdot ^\circ C) \) --- ### Step 1: Find the temperature change (\( \Delta T \)) \[ \Delta T = T_2 - T_1 = 75^\circ C - 12^\circ C = 63^\circ C \] **Reasoning:** Similarly, to find out how much energy is needed, determine how many degrees the water's temperature increases. --- ### Step 2: Calculate the heat energy required (\( Q \)) \[ Q = mc\Delta T \] Calculations: - \( 0.250\,kg \times 4186\,J/(kg \cdot ^\circ C) = 1046.5\,J/^\circ C \) - \( 1046.5 \times 63^\circ C \approx 65,919.5\,J \) **Result:** \[ Q \approx 65,920\,J \] --- ### Step 3: Find the heating time (\( t \)) \[ t = \frac{Q}{P} = \frac{65,920\,J}{80\,W} \approx 824\,s \] **Answer:** It will take approximately **824 seconds** (~13.7 minutes). --- ## Summary: - **First case:** About **350 seconds** to heat 120 g of water from 34°C to 73°C. - **Second case:** About **824 seconds** to heat 250 g of water from 12°C to 75°C. --- **Note:** All calculations assume perfect efficiency with no heat loss, as specified in the problem. If you'd like, I can also prepare a summarized version or explain any specific part in more detail!