Question

the initial temperature of a bomb calorimeter is 28.50°c. when a chemist carries out a reaction in this calorimeter, its temperature decreases to 27.45°c. if the calorimeter has a mass of 1.400 kg and a specific heat of 3.52 j/(g·°c), how much heat is absorbed by the reaction? use $q = mc_sdelta t$. 140 j 418 j 1,470 j 5,170 j

Explanation:

Step1: Convert mass to grams

Mass \( m = 1.400 \, \text{kg} = 1400 \, \text{g} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \))

Step2: Calculate temperature change

\( \Delta T = T_{\text{final}} - T_{\text{initial}} = 27.45^\circ\text{C} - 28.50^\circ\text{C} = -1.05^\circ\text{C} \)

Step3: Use the formula \( q = mC_p\Delta T \)

Given \( C_p = 3.52 \, \text{J/(g·°C)} \), substitute values:
\( q = 1400 \, \text{g} \times 3.52 \, \text{J/(g·°C)} \times (-1.05^\circ\text{C}) \)
The negative sign indicates heat is absorbed by the reaction (since the calorimeter loses heat, the reaction gains it). We can consider the magnitude for heat absorbed:
\( |q| = 1400 \times 3.52 \times 1.05 \)
\( 1400 \times 3.52 = 4928 \)
\( 4928 \times 1.05 = 5174.4 \approx 5170 \, \text{J} \) (rounded to a reasonable value matching the options)

Answer:

5,170 J (corresponding to the option with 5,170 J)