Let's solve each part step by step with explanations: ## **Given Data:** - Mass of ice, \( m_{ice} = 100 \) g - Initial temperature of ice = \(-40^\circ C\) - Mass of water added each time = 10g (a), 100g (b), 100g at 80°C (c) - Specific heat of ice, \( c_{ice} = 2.1 \) J/g°C - Specific heat of water, \( c_{water} = 4.18 \) J/g°C - Latent heat of fusion, \( L_f = 334 \) J/g - Assume insulated cup (no heat lost to surroundings). --- ## **a) 10g of \(10^\circ\)C water is added to the cup** ### **Step 1: Calculate energy required to bring ice from -40°C to 0°C** \[ q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T = 100 \times 2.1 \times (0 - (-40)) = 100 \times 2.1 \times 40 = 8400 \text{ J} \] ### **Step 2: Calculate energy that 10g of water at 10°C can provide to the system** \[ q_2 = m_{water} \cdot c_{water} \cdot \Delta T = 10 \times 4.18 \times (10 - 0) = 10 \times 4.18 \times 10 = 418 \text{ J} \] ### **Explanation & Conclusion:** - The 10g of water doesn't have enough energy to even warm the ice to 0°C. - Therefore, all water will freeze, and the system will be a mixture of ice and possibly some cold water at a temperature above -40°C but below 0°C. - **But since there is not enough energy, the final state:** All water freezes, temp is below 0°C. **Final Answers:** - **Temperature:** Below 0°C (exact value requires more calculation, but it’s < 0°C) - **Water in cup:** 0 g - **Ice in cup:** 110 g (100 g original ice + 10 g frozen water) --- ## **b) Another 100g of \(10^\circ\)C water is added to the cup** ### **Step 1: Total energy provided by 100g of 10°C water** \[ q_3 = 100 \times 4.18 \times (10 - 0) = 4180 \text{ J} \] ### **Step 2: Total energy required to warm all ice to 0°C (from part a, after first step, all at temp T < 0°C, total mass = 110g ice)** \[ \text{Let’s assume all ice is at initial temp (approximate):}\\ q_{ice} = 110 \times 2.1 \times (0 - T_{initial}) \] But let's do energy balance: - Energy to bring 110g ice from T to 0°C: If starting at -40°C, \[ q_{heat} = 110 \times 2.1 \times 40 = 9240 \text{ J} \] But we've already added 418 J before, so the ice is not at -40°C anymore. Let's approximate and focus on the key result: **Given the small amount of water, not all ice will melt.** ### **Step 3: Will all the ice melt?** - Total energy needed to melt all 110g of ice at 0°C: \[ q_{melt} = 110 \times 334 = 36740 \text{ J} \] The energy from water is far less than this, so not all ice melts. ### **Conclusion:** - The temperature will rise, but there will still be ice, and all water added will freeze or be cooled to 0°C. - **Final state:** All water freezes, and there is only ice at 0°C. **Final Answers:** - **Temperature:** 0°C - **Water in cup:** 0 g - **Ice in cup:** 110 g + 100 g = 210 g --- ## **c) Another 100g of water at 80°C is added** ### **Step 1: Energy provided by 100g of 80°C water cooling to 0°C** \[ q_4 = 100 \times 4.18 \times (80 - 0) = 100 \times 4.18 \times 80 = 33440 \text{ J} \] ### **Step 2: Remaining ice to melt** - Mass of ice before this step: 210g - Energy needed to melt all ice: \[ q_{melt,210g} = 210 \times 334 = 70140 \text{ J} \] But only 33440 J is available from the hot water, so only part of the ice will melt. \[ \text{Amount of ice melted} = \frac{33440 \text{ J}}{334 \text{ J/g}} = 100 \text{ g} \] ### **Step 3: Final mass calculation** - Initial ice: 210g - Ice melted: 100g - Ice left: 210g - 100g = 110g - Water formed: 100g (from melted ice) + 100g (added) = 200g **But, the melted ice is at 0°C, and the water added is also cooled to 0°C.** **Final Answers:** - **Temperature:** 0°C - **Water in cup:** 200 g - **Ice in cup:** 110 g --- ### **Summary Table** | Step | Temp (°C) | Water (g) | Ice (g) | Explanation | |------|-----------|-----------|---------|-------------| | a | < 0 | 0 | 110 | Not enough energy, all water freezes | | b | 0 | 0 | 210 | Still not enough energy, all water freezes, ice at 0°C | | c | 0 | 200 | 110 | 100g ice melts, rest is water at 0°C | Let me know if you'd like to see the **detailed calculations for the exact temperature in part (a)**!

Let's proceed step-by-step to find the exact temperature in part (a), considering energy exchanges. --- ### **Given Data Recap:** - Mass of ice, \( m_{ice} = 100\, \text{g} \) - Initial temperature of ice, \( T_{ice} = -40^\circ C \) - Mass of added water, \( m_{water} = 10\, \text{g} \) - Temperature of added water, \( T_{water} = 10^\circ C \) - Specific heat capacities: - Ice: \( c_{ice} = 2.1\, \text{J/g}^\circ C \) - Water: \( c_{water} = 4.18\, \text{J/g}^\circ C \) - Latent heat of fusion of ice: \( L_f = 334\, \text{J/g} \) --- ### **Step 1: Energy needed to warm the ice from \(-40^\circ C\) to \(0^\circ C\)** \[ q_{warm\,ice} = m_{ice} \times c_{ice} \times \Delta T = 100 \times 2.1 \times (0 - (-40)) = 100 \times 2.1 \times 40 = 8400\, \text{J} \] ### **Step 2: Energy released by the 10g of water cooling from \(10^\circ C\) to \(0^\circ C\)** \[ q_{cool\,water} = m_{water} \times c_{water} \times \Delta T = 10 \times 4.18 \times (10 - 0) = 10 \times 4.18 \times 10 = 418\, \text{J} \] ### **Step 3: Compare energy to warm ice to 0°C** - The water releases **418 J**, but warming the ice requires **8400 J**. - **Since 418 J < 8400 J**, the water is not enough to bring the ice to 0°C; thus, some of the ice will warm, but not all. --- ### **Step 4: Calculate the maximum temperature the ice can reach** The energy from the water is used first to warm the ice from \(-40^\circ C\) up to some temperature \(T_{ice}\) (which will be above \(-40^\circ C\) but below 0°C). The energy needed to warm the entire ice to \(T_{ice}\): \[ q_{warm\,ice} = m_{ice} \times c_{ice} \times (T_{ice} - (-40)) = 100 \times 2.1 \times (T_{ice} + 40) \] This must equal the energy supplied by the water: \[ 418 = 100 \times 2.1 \times (T_{ice} + 40) \] Solve for \(T_{ice}\): \[ T_{ice} + 40 = \frac{418}{100 \times 2.1} = \frac{418}{210} \approx 1.99 \] \[ T_{ice} \approx 1.99 - 40 = -38.01^\circ C \] **Interpretation:** - The ice warms from \(-40^\circ C\) to approximately **\(-38.01^\circ C\)** due to the energy released by the water. - **No melting occurs** because not enough energy is available to reach 0°C or melt any ice. --- ### **Step 5: Final temperature of the system** - The ice is at approximately \(-38^\circ C\). - The water is at \(0^\circ C\) (since it cools to the final equilibrium temperature \(T_f\), which is between \(-38^\circ C\) and \(0^\circ C\)). **But because the ice did not reach 0°C or melt, the system's temperature is effectively the temperature of the ice, i.e., approximately \(-38^\circ C\).** --- ### **Final Answer:** \[ \boxed{ \text{The temperature of the system, } x^\circ C \approx -38^\circ C } \] ### **Summary:** | Part | Final Temperature \(x^\circ C\) | Explanation | |--------|------------------------------|----------------------------------------------------| | a) | \(\approx -38^\circ C\) | Not enough energy to melt or warm ice to 0°C; ice warms slightly. | --- **Note:** The approximate temperature is just below 0°C, specifically about \(-38^\circ C\), based on energy calculations.