Question

how much energy does the water in this experiment absorb according to the calorimeter data?
calorimeter data
mass (g) | 100.0
specific heat (j/g°c) | 4.18
t_i (°c) | 21.2
t_f (°c) | 46.2
q_{h_2o} = ? j

Explanation:

Step1: Recall the formula for heat absorption

The formula to calculate 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

First, find \(\Delta T = T_f - T_i\). Substituting the given values: \(\Delta T = 46.2^\circ\text{C} - 21.2^\circ\text{C} = 25.0^\circ\text{C}\).

Step3: Substitute values into the heat formula

Now, substitute \(m = 100.0\ \text{g}\), \(c = 4.18\ \text{J/g}^\circ\text{C}\), and \(\Delta T = 25.0^\circ\text{C}\) into \(q = mc\Delta T\). So, \(q = 100.0\ \text{g} \times 4.18\ \text{J/g}^\circ\text{C} \times 25.0^\circ\text{C}\).

Step4: Perform the calculation

First, multiply \(100.0\) and \(4.18\): \(100.0 \times 4.18 = 418\). Then multiply by \(25.0\): \(418 \times 25.0 = 10450\ \text{J}\).

Answer:

\(10450\)