Question

a 50.0 g piece of granite was immersed in 28.3 g of water. the water and the calorimeter were initially 0.00 °c. the final temperature of the combination is 25.00 °c. what is the energy change of the water?
$c_{h_2o} = 4.18 \\, \text{j/g °c}$
$c_{cal} = 3.85 \\, \text{j/°c}$
$q_{h_2o} = ? \\, \text{j}$
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall 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.
For water, \(m = 28.3\space g\), \(c = 4.18\space J/g^\circ C\), and \(\Delta T = 25.00^\circ C - 0.00^\circ C = 25.00^\circ C\).

Step2: Calculate the heat change for water

Substitute the values into the formula:
\(q_{H_2O} = m_{H_2O} \times c_{H_2O} \times \Delta T\)
\(q_{H_2O} = 28.3\space g \times 4.18\space J/g^\circ C \times 25.00^\circ C\)
First, calculate \(28.3 \times 4.18\): \(28.3 \times 4.18 = 118.294\)
Then, multiply by 25.00: \(118.294 \times 25.00 = 2957.35\space J\)
Since the water is being heated (temperature increases), the energy change is positive.

Answer:

+2957 J (or more precisely +2957.35 J, depending on significant figures)