If 60.0 g of benzene at 25º C receives 1530 cal of heat, what will be its final temperature?

If 60.0 g of benzene at 25º C receives 1530 cal of heat, what will be its final temperature?

• A. 75.5º C
• B. 82.3º C
• C. 85.7º C
• D. 90.1º C

Answer: (C)

To determine the final temperature of benzene after absorbing heat, we can use the formula: \[ q = m \cdot C \cdot \Delta T \] where \( q \) is the heat absorbed (in calories), \( m \) is the mass of the substance (in grams), \( C \) is the specific heat capacity of the substance (for benzene, it is approximately \( 1.12 \, \text{cal/gºC} \)), and \( \Delta T \) is the change in temperature (final temperature minus initial temperature). Given: - Mass of benzene, \( m = 60.0 \, \text{g} \) - Heat absorbed, \( q = 1530 \, \text{cal} \) - Initial temperature, \( T_i = 25º C \) We need to find the final temperature \( T_f \). 1. Rearrange the formula to solve for \( \Delta T \): \[ \Delta T = \frac{q}{m \cdot C} \] 2. Plug in the values: \[ \Delta T = \frac{1530 \, \text{cal}}{60.0 \, \

To determine the final temperature of benzene after absorbing heat, we can use the formula:

[ q = m \cdot C \cdot \Delta T ]

where ( q ) is the heat absorbed (in calories), ( m ) is the mass of the substance (in grams), ( C ) is the specific heat capacity of the substance (for benzene, it is approximately ( 1.12 , \text{cal/gºC} )), and ( \Delta T ) is the change in temperature (final temperature minus initial temperature).

Given:

• Mass of benzene, ( m = 60.0 , \text{g} )

• Heat absorbed, ( q = 1530 , \text{cal} )

• Initial temperature, ( T_i = 25º C )

We need to find the final temperature ( T_f ).

1. Rearrange the formula to solve for ( \Delta T ):

[ \Delta T = \frac{q}{m \cdot C} ]

2. Plug in the values:

[ \Delta T = \frac{1530 , \text{cal}}{60.0 , \