Chemistry Regents Practice Test 2026 – The All-in-One Guide to Exam Success!

Question: 1 / 400

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?

28°C

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

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32°C

34°C

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