Zinc metal has a specific heat capacity of 0.388????°??. ?If 5.26?? ?of Zinc, with a temperature of 60°??, ?was placed in a calorimeter with 50.0 ?grams of water. If the final temperature of the water was 27.5°??, ?what will be the initial temperature of the water in the calorimeter?
The Correct Answer and Explanation is :
Problem:
Zinc metal has a specific heat capacity of 0.388 J/g°C. If 5.26 g of Zinc at 60°C is placed in a calorimeter containing 50.0 g of water, and the final temperature of the water is 27.5°C, we are asked to find the initial temperature of the water in the calorimeter.
Solution:
To find the initial temperature of the water, we can use the principle of conservation of energy, where the heat lost by the zinc metal equals the heat gained by the water in the calorimeter.
- Heat Lost by Zinc:
The formula for heat lost or gained is:
[
Q = mc\Delta T
]
where:
- ( Q ) is the heat energy (in Joules),
- ( m ) is the mass (in grams),
- ( c ) is the specific heat capacity (in J/g°C),
- ( \Delta T ) is the change in temperature (in °C).
For zinc:
- Mass of zinc = 5.26 g,
- Specific heat capacity of zinc = 0.388 J/g°C,
- Initial temperature of zinc = 60°C,
- Final temperature of zinc = 27.5°C (same as the final temperature of water).
The change in temperature for zinc is:
[
\Delta T_{\text{zinc}} = 60°C – 27.5°C = 32.5°C
]
Now, calculate the heat lost by zinc:
[
Q_{\text{zinc}} = (5.26 \, \text{g}) \times (0.388 \, \text{J/g°C}) \times (32.5°C) = 66.56 \, \text{J}
]
- Heat Gained by Water:
Let the initial temperature of the water be ( T_{\text{initial}} ). The final temperature of water is given as 27.5°C.
The heat gained by the water can be calculated using the same formula:
[
Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}}
]
where:
- Mass of water = 50.0 g,
- Specific heat capacity of water = 4.18 J/g°C,
- ( \Delta T_{\text{water}} = 27.5°C – T_{\text{initial}} ).
The heat gained by the water must be equal to the heat lost by the zinc, so:
[
Q_{\text{water}} = 66.56 \, \text{J}
]
Thus:
[
50.0 \, \text{g} \times 4.18 \, \text{J/g°C} \times (27.5°C – T_{\text{initial}}) = 66.56 \, \text{J}
]
Now solve for ( T_{\text{initial}} ):
[
209 \times (27.5 – T_{\text{initial}}) = 66.56
]
[
27.5 – T_{\text{initial}} = \frac{66.56}{209}
]
[
27.5 – T_{\text{initial}} = 0.318
]
[
T_{\text{initial}} = 27.5 – 0.318 = 27.18°C
]
Final Answer:
The initial temperature of the water in the calorimeter was approximately 27.18°C.
Explanation:
The key to solving this problem is recognizing the principle of conservation of energy, which states that the heat lost by the hot object (zinc) equals the heat gained by the cold object (water) in a closed system. By applying the formula for heat transfer and setting the heat lost by zinc equal to the heat gained by water, we can calculate the initial temperature of the water.
This calculation assumes no heat loss to the environment, which is why we can equate the energy transfers. The zinc metal cools down as it transfers heat to the water, which causes the water’s temperature to rise until both substances reach thermal equilibrium at 27.5°C.