Chemistry Regents Practice Test

A 36-gram sample of water has an initial temperature of 22°C. After the sample absorbs 1200 joules of heat energy, the final temperature of the sample is?

1. 28°C

2. 30°C

3. 32°C

4. 34°C

The correct answer is: 30°C

To determine the final temperature of the water after absorbing 1200 joules of heat energy, we can use the specific heat formula, which is: \[ q = mc\Delta T \] where: - \( q \) is the amount of heat absorbed (in joules), - \( m \) is the mass of the substance (in grams), - \( c \) is the specific heat capacity (for water, it's approximately \( 4.18 \, \text{J/g°C} \)), - \( \Delta T \) is the change in temperature (final temperature - initial temperature). In this scenario, we are given: - \( m = 36 \, \text{g} \), - \( q = 1200 \, \text{J} \), - The initial temperature (\( T_i \)) is \( 22°C \). First, we can rearrange the formula to solve for the change in temperature (\( \Delta T \)): \[ \Delta T = \frac{q}{mc} \] Substituting the values into this equation gives us: \[ \Delta T = \frac{1200 \, \text{J}}{36 \, \text