Question

how much energy does the water in this experiment absorb according to the calorimeter data? calorimeter data mass (g) 300.0 specific heat (j/g°c) 4.18 t_i (°c) 20.0 t_f (°c) 36.0 q_{h_2o} = ? j

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed or released is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T = T_f - T_i \).

Step2: Calculate temperature change

\( \Delta T = 36.0 - 20.0 = 16.0^\circ\text{C} \)

Step3: Substitute values into formula

\( m = 300.0\ \text{g} \), \( c = 4.18\ \text{J/g}^\circ\text{C} \), \( \Delta T = 16.0^\circ\text{C} \)
\( q = 300.0 \times 4.18 \times 16.0 \)
\( q = 300.0 \times 66.88 \)
\( q = 20064\ \text{J} \)

Answer:

20064