Question

the heat capacity of the calorimeter vessel is 420 j/°c. the temperature changed from 20.00 °c to 22.14 °c. calculate the quantity of energy (kj) transferred to the 600 g water in the bath. specific heat capacity h₂o = 4.18×10⁻³ kj g⁻¹ °c⁻¹. δt = 2.14 °c

Explanation:

Step1: Calculate heat absorbed by water

Use the formula $Q = mc\Delta T$, where $m = 600\ g$, $c=4.18\times 10^{-3}\ kJg^{-1}{^{\circ}C}^{-1}$, and $\Delta T = 2.14^{\circ}C$.
$Q_{water}=600\times4.18\times 10^{-3}\times2.14$
$Q_{water}=600\times4.18\times2.14\times 10^{-3}$
$Q_{water}=5.35\ kJ$

Step2: Calculate heat absorbed by calorimeter

Use the formula $Q = C\Delta T$, where $C = 420\ J/^{\circ}C=0.42\ kJ/^{\circ}C$ and $\Delta T = 2.14^{\circ}C$.
$Q_{calorimeter}=0.42\times2.14 = 0.8988\ kJ$

Step3: Calculate total energy transferred

Total energy $Q_{total}=Q_{water}+Q_{calorimeter}$
$Q_{total}=5.35 + 0.8988=6.2488\ kJ\approx6.25\ kJ$

Answer:

$6.25\ kJ$