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
tᵢ (°c) | 21.2
tբ (°c) | 46.2
qₕ₂ₒ = ? 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 mass, \(c\) is specific heat, and \(\Delta T\) is the change in temperature (\(T_f - T_i\)).

Step2: Calculate the change in temperature

\(\Delta T = T_f - T_i = 46.2^\circ C - 21.2^\circ C = 25^\circ C\)

Step3: Substitute values into the formula

Given \(m = 100.0\ g\), \(c = 4.18\ J/g^\circ C\), and \(\Delta T = 25^\circ C\), we have:
\(q = 100.0\ g \times 4.18\ J/g^\circ C \times 25^\circ C\)

Step4: Perform the calculation

First, multiply \(100.0\) and \(4.18\): \(100.0 \times 4.18 = 418\)
Then, multiply by \(25\): \(418 \times 25 = 10450\)

Answer:

\(10450\)