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 calorimeter?
$c_{h_2o} = 4.184 \\, \text{j/g °c}$
$c_{cal} = 3.86 \\, \text{j/°c}$
$q_{cal} = \\,?\\, \\, \text{j}$
enter either a + or - sign and the magnitude.

Explanation:

Step1: Identify the formula for heat change of calorimeter

The formula for the heat change of the calorimeter (\(q_{cal}\)) is \(q_{cal}=c_{cal}\times\Delta T\), where \(c_{cal}\) is the specific heat capacity of the calorimeter and \(\Delta T\) is the change in temperature.

Step2: Calculate the change in temperature (\(\Delta T\))

The initial temperature (\(T_i\)) is \(85.0\ ^{\circ}\text{C}\) and the final temperature (\(T_f\)) is \(52.0\ ^{\circ}\text{C}\). So, \(\Delta T = T_f - T_i=52.0 - 85.0=- 33.0\ ^{\circ}\text{C}\)

Step3: Calculate \(q_{cal}\)

We know that \(c_{cal} = 3.86\ \text{J}/^{\circ}\text{C}\) and \(\Delta T=-33.0\ ^{\circ}\text{C}\). Substitute these values into the formula \(q_{cal}=c_{cal}\times\Delta T\)

\(q_{cal}=3.86\ \text{J}/^{\circ}\text{C}\times(- 33.0\ ^{\circ}\text{C})\)

\(q_{cal}=- 127.38\ \text{J}\approx - 127\ \text{J}\) (or more precisely, if we consider significant figures, since the values have three significant figures in temperature change and \(c_{cal}\) has three, the result should be \(- 127\ \text{J}\) or \(-1.27\times 10^{2}\ \text{J}\))

Wait, let's recalculate: \(3.86\times(-33)=3.86\times(-30 - 3)=-115.8-11.58=-127.38\approx - 127\ \text{J}\) (or - 127 J when rounded to three significant figures)

Answer:

\(-127\) (or \(-1.27\times 10^{2}\))