A `1.0g ` sample of substance `A` at `100^(@)C` is added to `100 mL` of `H_(2)O` at `25^(@)C`. Using separate `100 mL` portions of `H_(2)O`, the procedure is repeated with substance `B` and then with substance `C`. How will the final temperatures of the water compare ?
`{:("Substance","Specific heat"),(A,0.60 J g^(-1) "^(@)C^(-1)),((B,0.40 J g^(-1) "^(@)C^(-1))),(C,0.20 J g^(-1) "^(@)C^(-1)):}`

Options

A. `T_C gt T_B gtT_A`
B. `T_BgtT_A gtT_C`
C. `T_A gtT_BgtT_C`
D. `T_A=T_B=T_C`

Answer

To solve the problem, we need to understand how the specific heat of the substances affects the final temperature of the water when each substance is added. The specific heat capacity indicates how much heat energy is required to raise the temperature of a substance.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Substance A: 1.0 g at 100°C
- Water: 100 mL at 25°C
- Specific heat of A = 0.60 J/g°C
- Specific heat of B = 0.40 J/g°C
- Specific heat of C = 0.20 J/g°C

2. **Understand the Heat Transfer**:
When a hot substance is added to cooler water, heat will transfer from the substance to the water until thermal equilibrium is reached. The heat lost by the substance will be equal to the heat gained by the water.

3. **Calculate the Heat Transfer**:
The heat lost by the substance can be calculated using the formula:
\[ q = m \cdot c \cdot \Delta T \]
where \( q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature.

4. **Set Up the Equations for Each Substance**:
For each substance, we can set up the equation for heat transfer:
- For substance A:
\[ q_A = 1.0 \, \text{g} \cdot 0.60 \, \text{J/g°C} \cdot (100 - T_A) \]
- For substance B:
\[ q_B = 1.0 \, \text{g} \cdot 0.40 \, \text{J/g°C} \cdot (100 - T_B) \]
- For substance C:
\[ q_C = 1.0 \, \text{g} \cdot 0.20 \, \text{J/g°C} \cdot (100 - T_C) \]

5. **Heat Gained by Water**:
The heat gained by the water can be calculated as:
\[ q_{water} = 100 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (T_f - 25) \]
where \( T_f \) is the final temperature of the water.

6. **Equate Heat Lost and Gained**:
For each substance, set the heat lost equal to the heat gained by the water:
- For A:
\[ 1.0 \cdot 0.60 \cdot (100 - T_A) = 100 \cdot 4.18 \cdot (T_f - 25) \]
- For B:
\[ 1.0 \cdot 0.40 \cdot (100 - T_B) = 100 \cdot 4.18 \cdot (T_f - 25) \]
- For C:
\[ 1.0 \cdot 0.20 \cdot (100 - T_C) = 100 \cdot 4.18 \cdot (T_f - 25) \]

7. **Analyze the Final Temperatures**:
Since the specific heat capacities are different, the substance with the highest specific heat (substance A) will result in the highest final temperature of the water. Conversely, the substance with the lowest specific heat (substance C) will result in the lowest final temperature of the water.

8. **Conclusion**:
Therefore, the final temperatures can be compared as:
\[ T_A > T_B > T_C \]

### Final Answer:
The final temperatures of the water will be in the order:
\[ T_A > T_B > T_C \]