Question

a 25.5 g piece of metal with a temperature of 105.00 °c is dropped into a calorimeter containing 145 g water. the water temperature increases from 20.00 °c to 23.50 °c. the heat capacity of the calorimeter is 6.2 j/°c and the specific heat capacity of water is 4.18 j/g °c. what is the energy change for the water? q_{h_2o} = ? j enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for heat energy

The formula to calculate the heat energy (\(q\)) absorbed or released by a substance is \(q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. For water, since the temperature increases, it absorbs heat, so the sign will be positive.

Step2: Identify the values for water

• Mass of water (\(m\)) = 145 g
• Specific heat capacity of water (\(c\)) = \(4.18 \, \text{J/g}^\circ\text{C}\)
• Initial temperature (\(T_i\)) = \(20.00^\circ\text{C}\)
• Final temperature (\(T_f\)) = \(23.50^\circ\text{C}\)
• Change in temperature (\(\Delta T\)) = \(T_f - T_i = 23.50 - 20.00 = 3.50^\circ\text{C}\)

Step3: Calculate the heat energy for water

Substitute the values into the formula:
\(q_{\text{H}_2\text{O}} = m \times c \times \Delta T\)
\(q_{\text{H}_2\text{O}} = 145 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 3.50^\circ\text{C}\)
First, calculate \(145 \times 4.18 = 606.1\)
Then, \(606.1 \times 3.50 = 2121.35\)

Answer:

\(+2121.35\)