A 150 g copper bowl contains 260 g of water, both at `20.0^(@)C` A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.00 g being converted to steam. The final temperature of the system is `100^(@)C`. Neglect energy transfers with the environment. (a) How much energy ( in calories ) is transferred to the water as heat ? (b) How much to the bowl ? ( c ) What is the original temperature of the cylinder ?

Answer

To solve the problem step by step, we will break it down into three parts as per the question.

### Part (a): Energy Transferred to the Water

1. **Calculate the energy required to heat the water from 20 °C to 100 °C.**
- Mass of water (m_water) = 260 g
- Specific heat of water (c_water) = 1 cal/g°C
- Temperature change (ΔT_water) = 100 °C - 20 °C = 80 °C
- Energy (Q_water) = m_water × c_water × ΔT_water
\[ Q_{\text{water}} = 260 \, \text{g} \times 1 \, \text{cal/g°C} \times 80 \, \text{°C} = 20800 \, \text{cal} \]

2. **Calculate the energy required to convert 5 g of water to steam.**
- Mass of steam (m_steam) = 5 g
- Latent heat of vaporization of water (L) = 539 cal/g
- Energy for vaporization (Q_vaporization) = m_steam × L
\[ Q_{\text{vaporization}} = 5 \, \text{g} \times 539 \, \text{cal/g} = 2695 \, \text{cal} \]

3. **Total energy transferred to the water.**
\[ Q_{\text{total}} = Q_{\text{water}} + Q_{\text{vaporization}} = 20800 \, \text{cal} + 2695 \, \text{cal} = 23495 \, \text{cal} \]

### Part (b): Energy Transferred to the Bowl

1. **Calculate the energy required to heat the copper bowl from 20 °C to 100 °C.**
- Mass of the bowl (m_bowl) = 150 g
- Specific heat of copper (c_copper) = 0.0923 cal/g°C
- Temperature change (ΔT_bowl) = 100 °C - 20 °C = 80 °C
- Energy (Q_bowl) = m_bowl × c_copper × ΔT_bowl
\[ Q_{\text{bowl}} = 150 \, \text{g} \times 0.0923 \, \text{cal/g°C} \times 80 \, \text{°C} = 1100 \, \text{cal} \]

### Part (c): Original Temperature of the Cylinder

1. **Calculate the total heat gained by the water and the bowl.**
\[ Q_{\text{gained}} = Q_{\text{total}} + Q_{\text{bowl}} = 23495 \, \text{cal} + 1100 \, \text{cal} = 24595 \, \text{cal} \]

2. **Set this equal to the heat lost by the cylinder.**
- Mass of the cylinder (m_cylinder) = 300 g
- Specific heat of the cylinder (c_cylinder) = 0.0923 cal/g°C
- Let the original temperature of the cylinder be T_initial.
- Heat lost by the cylinder (Q_lost) = m_cylinder × c_cylinder × (T_initial - 100)
\[ Q_{\text{lost}} = 300 \, \text{g} \times 0.0923 \, \text{cal/g°C} \times (T_{\text{initial}} - 100) = 24595 \, \text{cal} \]

3. **Solve for T_initial.**
\[ 300 \times 0.0923 \times (T_{\text{initial}} - 100) = 24595 \]
\[ 27.69 \times (T_{\text{initial}} - 100) = 24595 \]
\[ T_{\text{initial}} - 100 = \frac{24595}{27.69} \approx 888.4 \]
\[ T_{\text{initial}} \approx 988.4 \, \text{°C} \]

### Final Answers:
(a) Energy transferred to the water: **23495 cal**
(b) Energy transferred to the bowl: **1100 cal**
(c) Original temperature of the cylinder: **988.4 °C**