Question

constant - volume calorimeters are sometimes calibrated by running a combustion reaction of known and measuring the change in temperature. for example, the combustion energy of glucose is 15.57 kj/g. when a 7.7651 - g sample of glucose burns in a constant - volume calorimeter, the calorimeter temperature increases from 16.73°c to 42.74°c. find the total heat capacity of the calorimeter.

Explanation:

Step1: Calculate heat released by glucose

The heat released ($q$) by the glucose combustion is calculated using the mass ($m$) of glucose and its combustion - energy ($\Delta H_{comb}$). The formula is $q = m\times\Delta H_{comb}$. Given $m = 7.7651\ g$ and $\Delta H_{comb}=15.57\ kJ/g$, so $q=7.7651\ g\times15.57\ kJ/g = 7.7651\times15.57\ kJ=120.903607\ kJ$.

Step2: Calculate the temperature change

The temperature change ($\Delta T$) is calculated as $\Delta T=T_{final}-T_{initial}$. Given $T_{initial}=16.73^{\circ}C$ and $T_{final}=42.74^{\circ}C$, so $\Delta T = 42.74^{\circ}C - 16.73^{\circ}C=26.01^{\circ}C$.

Step3: Calculate the heat - capacity of the calorimeter

The heat - capacity ($C$) of the calorimeter is calculated using the formula $C=\frac{q}{\Delta T}$. Substituting $q = 120.903607\ kJ$ and $\Delta T = 26.01^{\circ}C$ into the formula, we get $C=\frac{120.903607\ kJ}{26.01^{\circ}C}\approx4.65\ kJ/^{\circ}C$.

Answer:

$4.65$