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_pdelta t ).

○ 140 j
○ 418 j
○ 1,470 j
○ 5,170 j

Explanation:

Step1: Convert mass to grams

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

Step2: Calculate temperature change

\( \Delta T = T_{\text{final}} - T_{\text{initial}} = 27.45^\circ\text{C} - 28.50^\circ\text{C} = -1.05^\circ\text{C} \) (But we take absolute value for heat calculation as we are interested in magnitude of heat absorbed by reaction, and the negative sign indicates heat loss from calorimeter, so reaction absorbs heat equal to heat lost by calorimeter)

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

Substitute \( m = 1400 \, \text{g} \), \( c_p = 3.52 \, \text{J/(g}\cdot^\circ\text{C)} \), \( \Delta T = 1.05^\circ\text{C} \) (absolute value)
\( q = 1400 \times 3.52 \times 1.05 \)
First, \( 1400 \times 3.52 = 4928 \)
Then, \( 4928 \times 1.05 = 4928 + 4928\times0.05 = 4928 + 246.4 = 5174.4 \approx 5170 \, \text{J} \) (rounded to match options)

Answer:

5,170 J (The option with 5,170 J, assuming the last option is D. 5,170 J (if options are labeled A to D as per order: A. 140 J, B. 418 J, C. 1,470 J, D. 5,170 J then D. 5,170 J))