`20.0 g` of ammonium nitrate `(NH_(4)NO_(3))` is dissolved In `125 g` of water in a coffee-cup calorimeter, the temperature falls from `296.5 K` to `286.4 K`. Find the value of `q` for the calorimeter. (Hint: heat capacity of water as the heat capacity of the calorimeter and its content)

Explanation

To find the value of \( q \) for the calorimeter when \( 20.0 \, \text{g} \) of ammonium nitrate \( (NH_4NO_3) \) is dissolved in \( 125 \, \text{g} \) of water and the temperature falls from \( 296.5 \, \text{K} \) to \( 286.4 \, \text{K} \), we can follow these steps:

### Step 1: Identify the known values
- Mass of water, \( m = 125 \, \text{g} \)
- Specific heat capacity of water, \( C = 4.18 \, \text{J/g°C} \)
- Initial temperature, \( T_i = 296.5 \, \text{K} \)
- Final temperature, \( T_f = 286.4 \, \text{K} \)

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