Question

in the laboratory a \coffee cup\ calorimeter, or constant pressure calorimeter, is frequently used to determine the heat of a solid, or to measure the energy of a solution phase reaction. a student heats 64.15 grams of iron to 98.55 °c and then drops it into a cup containing 81.31 grams of water at 23.91 °c. she measures the final temperature to be 29.85 °c. the heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.90 j/°c. assuming that no heat is lost to the surroundings calculate the specific heat of iron. specific heat (fe) = j/g °c. show approach show tutor steps

Explanation:

Step1: Recall heat - transfer formula

The heat lost by the iron is equal to the heat gained by the water and the calorimeter. The formula for heat transfer is $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature. Let $c_{Fe}$ be the specific heat of iron, $m_{Fe}=64.15\ g$ be the mass of iron, $\Delta T_{Fe}=98.55 - 29.85=68.7^{\circ}C$ be the change in temperature of iron, $m_{w}=81.31\ g$ be the mass of water, $c_{w} = 4.184\ J/g^{\circ}C$ be the specific heat of water, $\Delta T_{w}=29.85 - 23.91 = 5.94^{\circ}C$ be the change in temperature of water, and $C_{cal}=1.90\ J/^{\circ}C$ be the heat - capacity of the calorimeter.

Step2: Set up heat - balance equation

$q_{lost,Fe}=q_{gained,w}+q_{gained,cal}$. Using the heat - transfer formula, we have $m_{Fe}c_{Fe}\Delta T_{Fe}=m_{w}c_{w}\Delta T_{w}+C_{cal}\Delta T_{cal}$. Since $\Delta T_{cal}=\Delta T_{w}=5.94^{\circ}C$, we substitute the known values: $64.15\ g\times c_{Fe}\times68.7^{\circ}C=81.31\ g\times4.184\ J/g^{\circ}C\times5.94^{\circ}C + 1.90\ J/^{\circ}C\times5.94^{\circ}C$.

Step3: Calculate the right - hand side of the equation

First, calculate $m_{w}c_{w}\Delta T_{w}$: $81.31\ g\times4.184\ J/g^{\circ}C\times5.94^{\circ}C=81.31\times4.184\times5.94\ J\approx2027.9\ J$. Then calculate $C_{cal}\Delta T_{cal}$: $1.90\ J/^{\circ}C\times5.94^{\circ}C = 11.29\ J$. The sum of these two values is $2027.9\ J+11.29\ J=2039.19\ J$.

Step4: Solve for $c_{Fe}$

We have the equation $64.15\ g\times c_{Fe}\times68.7^{\circ}C=2039.19\ J$. Then $c_{Fe}=\frac{2039.19\ J}{64.15\ g\times68.7^{\circ}C}$. $64.15\times68.7 = 64.15\times(70 - 1.3)=64.15\times70-64.15\times1.3=4490.5-83.4=4407.1$. So $c_{Fe}=\frac{2039.19}{4407.1}\ J/g^{\circ}C\approx0.463\ J/g^{\circ}C$.

Answer:

$0.463$