Question

1. given that when, in the reaction above, 2.000 g of ch4(g) is used in a calorimeter with heat capacity of 2139 j/°c. the temperature falls from 25.26 to 22.88 °c.

Explanation:

Step1: Calculate the temperature change

$\Delta T=T_2 - T_1=22.88^{\circ}C - 25.26^{\circ}C=- 2.38^{\circ}C$

Step2: Use the heat - capacity formula

The heat absorbed or released by the calorimeter is given by $q = C\times\Delta T$, where $C = 2139\ J/^{\circ}C$ and $\Delta T=-2.38^{\circ}C$.
$q=2139\ J/^{\circ}C\times(- 2.38^{\circ}C)=-5090.82\ J$
The negative sign indicates that heat is released by the system.

Answer:

$5090.82\ J$