Question

a 245.0 g sample of lithium was placed in a calorimeter containing 45.0 g of water. the water and the calorimeter were initially 85.0 °c. the system reaches equilibrium at 52.0 °c. what is the energy change for the water?
$c_{h_2o} = 4.184$ j/g °c
$c_{cal} = 3.86$ j/°c
$q_{h_2o} = ?$ j
enter either a + or - sign and the magnitude.

Explanation:

Step1: Identify the formula for heat change

The formula for heat change (\(q\)) is \(q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature.

Step2: Determine the values

For water, \(m = 45.0\space g\), \(c = 4.184\space J/g^\circ C\), initial temperature \(T_i = 85.0^\circ C\), final temperature \(T_f = 52.0^\circ C\).
\(\Delta T = T_f - T_i = 52.0 - 85.0 = -33.0^\circ C\)

Step3: Calculate the heat change for water

Substitute the values into the formula:
\(q_{H_2O} = 45.0\space g\times4.184\space J/g^\circ C\times(-33.0^\circ C)\)
First, calculate \(45.0\times4.184 = 188.28\)
Then, \(188.28\times(-33.0) = -6213.24\space J\)

Answer:

\(-6213.24\)