Question

complete combustion of a 0.350 g sample of a compound in a bomb calorimeter releases 14.0 kj of heat. the bomb calorimeter has a mass of 1.20 kg and a specific heat of 3.55 j/(g·°c). if the initial temperature of the calorimeter is 22.5°c, what is its final temperature? use $q = mc_pdelta t$. \\(\circ\\) 19.2°c \\(\circ\\) 25.8°c \\(\circ\\) 34.2°c \\(\circ\\) 72.3°c

Explanation:

Step1: Convert units

The heat released \( q = 14.0\space kJ=14000\space J \), mass of calorimeter \( m = 1.20\space kg = 1200\space g \), specific heat \( C_p=3.55\space J/(g\cdot^\circ C) \), initial temperature \( T_i = 22.5^\circ C \).

Step2: Rearrange formula for \(\Delta T\)

From \( q = mC_p\Delta T \), we get \(\Delta T=\frac{q}{mC_p}\).

Step3: Calculate \(\Delta T\)

Substitute values: \(\Delta T=\frac{14000\space J}{1200\space g\times3.55\space J/(g\cdot^\circ C)}\approx 3.3^\circ C\).

Step4: Find final temperature \( T_f \)

\( T_f=T_i + \Delta T=22.5^\circ C+ 3.3^\circ C = 25.8^\circ C \).

Answer:

25.8°C (corresponding to the option: 25.8°C)