A solid of mass 100 g at a temperature of 90°C is placed in 100g of water at 20°C in a container of negligible heat capacity. If the final steady temperature is 60°C, calculate the... A solid of mass 100 g at a temperature of 90°C is placed in 100g of water at 20°C in a container of negligible heat capacity. If the final steady temperature is 60°C, calculate the specific heat capacity of the solid, given that the specific heat capacity of water equals to 4.2x10³ jkg⁻¹K⁻¹.
Understand the Problem
The question describes a calorimetry experiment where a solid object at a higher temperature is placed in water at a lower temperature. The goal is to determine the specific heat capacity of the solid using the principle of heat exchange and the given information about the masses, temperatures, and specific heat capacity of water.
Answer
The specific heat capacity of the metal is $0.507 \, \text{J/g°C}$.
Answer for screen readers
The specific heat capacity of the metal is $0.507 , \text{J/g°C}$.
Steps to Solve
- Calculate the heat gained by the water
We'll use the formula $Q = mc\Delta T$, where $Q$ is heat, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.
For the water:
$m_{water} = 100.0 , \text{g}$ $c_{water} = 4.184 , \text{J/g°C}$ $\Delta T_{water} = 25.6°\text{C} - 22.0°\text{C} = 3.6°\text{C}$
$Q_{water} = (100.0 , \text{g}) \times (4.184 , \text{J/g°C}) \times (3.6°\text{C})$ $Q_{water} = 1506.24 , \text{J}$
- Calculate the heat lost by the metal
The heat lost by the metal is equal to the negative of the heat gained by the water since the calorimeter is perfectly insulated.
$Q_{metal} = -Q_{water} = -1506.24 , \text{J}$
- Determine the specific heat capacity of the metal
We rearrange the heat equation to solve for the specific heat capacity of the metal: $c = \frac{Q}{m\Delta T}$
For the metal:
$m_{metal} = 50.0 , \text{g}$ $Q_{metal} = -1506.24 , \text{J}$ $\Delta T_{metal} = 25.6°\text{C} - 85.0°\text{C} = -59.4°\text{C}$
$c_{metal} = \frac{-1506.24 , \text{J}}{(50.0 , \text{g}) \times (-59.4°\text{C})}$ $c_{metal} = \frac{-1506.24 , \text{J}}{-2970 , \text{g°C}}$ $c_{metal} = 0.507 , \text{J/g°C}$
The specific heat capacity of the metal is $0.507 , \text{J/g°C}$.
More Information
The specific heat capacity is a material property that describes the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius. This value is useful in identifying the type of metal used in the experiment. For example, the specific heat capacity for Iron is $0.450 , \text{J/g°C}$, and for Copper is $0.385 , \text{J/g°C}$. The calculated specific heat capacity $0.507 , \text{J/g°C}$ does not match either, suggesting the metal is likely Aluminum at $0.900 , \text{J/g°C}$ or another unique alloy.
Tips
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Incorrect sign for heat transfer: Forgetting that the metal loses heat (negative Q) while the water gains heat (positive Q). Always consider the direction of heat flow.
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Incorrect temperature difference: Calculating $\Delta T$ as initial minus final instead of final minus initial. This will lead to a sign error and an incorrect specific heat capacity.
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Units: Forgetting units for each value will lead to not being able to cancel out the units properly.
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