Question

how much energy does the water in this experiment absorb according to the calorimeter data?
calorimeter data
mass (g) | 50.0
specific heat (j/g°c) | 4.18
ti (°c) | 20.5
tf (°c) | 32.5
q_{h2o} = ? j

Explanation:

Step1: Recall the formula for heat absorption

The formula for the heat absorbed (\(q\)) by a substance is \(q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature (\(T_f - T_i\)).

Step2: Calculate the change in temperature

\(\Delta T = T_f - T_i = 32.5^\circ\text{C} - 20.5^\circ\text{C} = 12.0^\circ\text{C}\)

Step3: Substitute values into the heat formula

We know \(m = 50.0\ \text{g}\), \(c = 4.18\ \text{J/g}^\circ\text{C}\), and \(\Delta T = 12.0^\circ\text{C}\). Plugging these into \(q = mc\Delta T\):

\(q = 50.0\ \text{g} \times 4.18\ \text{J/g}^\circ\text{C} \times 12.0^\circ\text{C}\)

First, multiply \(50.0\) and \(4.18\): \(50.0 \times 4.18 = 209\)

Then, multiply the result by \(12.0\): \(209 \times 12.0 = 2508\)

Answer:

\(2508\)