A solid of mass 100 g at a temperature of 90°C is placed in 100 g of water at 20°C in a container of negligible heat capacity. If the final steady temperature is 60°C, calculate th... A solid of mass 100 g at a temperature of 90°C is placed in 100 g 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. The specific heat capacity of water is 4.2 x 10³ J/kg·K.
Understand the Problem
The question describes a calorimetry experiment where a solid at a higher temperature is placed in water at a lower temperature. You're asked to calculate the specific heat capacity of the solid using the principle of heat exchange, where the heat lost by the solid equals the heat gained by the water. The final steady temperature and the specific heat capacity of water are given.
Answer
$0.472 \, \text{J/g}^\circ\text{C}$
Answer for screen readers
$c_{solid} = 0.472 , \text{J/g}^\circ\text{C}$
Steps to Solve
- Calculate the heat gained by the water
The heat gained by the water can be calculated using the formula $Q = mc\Delta{T}$, where $Q$ is the heat, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta{T}$ is the change in temperature. The mass of the water is 100.0 g, the specific heat capacity of water is 4.184 J/g°C, and the change in temperature is (25.1°C - 22.0°C) = 3.1°C.
$Q_{water} = (100.0 , \text{g}) \times (4.184 , \text{J/g}^\circ\text{C}) \times (3.1 , ^\circ\text{C})$ $Q_{water} = 1297.04 , \text{J}$
- Calculate the heat lost by the solid
The heat lost by the solid is equal to the negative of the heat gained by the water, assuming no heat is lost to the surroundings. Therefore, $Q_{solid} = -Q_{water}$. So, $Q_{solid} = -1297.04 , \text{J}$.
- Calculate the specific heat capacity of the solid
We use the formula $Q = mc\Delta{T}$ again, but this time we solve for $c$, the specific heat capacity of the solid: $c = \frac{Q}{m\Delta{T}}$. The mass of the solid is 50.0 g, the heat lost by the solid is -1297.04 J, and the change in temperature of the solid is (25.1°C - 80.0°C) = -54.9°C.
$c_{solid} = \frac{-1297.04 , \text{J}}{(50.0 , \text{g}) \times (-54.9 , ^\circ\text{C})}$ $c_{solid} = \frac{-1297.04 , \text{J}}{-2745 , \text{g}^\circ\text{C}}$ $c_{solid} = 0.4724 , \text{J/g}^\circ\text{C}$
$c_{solid} = 0.472 , \text{J/g}^\circ\text{C}$
More Information
The specific heat capacity is an intensive property that indicates the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius. It is a characteristic property and is different for different substances.
Tips
A common mistake is forgetting to include the negative sign when calculating the heat lost by the solid. Also, students might mix up the temperature changes for the water and the solid. Always ensure you are subtracting the initial temperature from the final temperature $(\Delta{T} = T_{final} - T_{initial})$. Another common mistake involves using incorrect units. Make sure to keep track of the units throughout the calculation and ensure they cancel correctly.
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