Question

1. 2500 grams of 1-propanol solution was used in a calorimeter as the heat absorbing substance. 143.75 kj of heat caused the 1-propanol to undergo a temperature change from 10.0 °c to 35.0 °c. calculate the specific heat capacity for the 1-propanol. (answer: 2.3 kj/kg°c)

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed \( Q \) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. We need to solve for \( c \), so rearrange the formula to \( c=\frac{Q}{m\Delta T} \).

Step2: Convert mass to kg

The mass \( m = 2500 \) grams. Since \( 1 \) kg \( = 1000 \) grams, \( m=\frac{2500}{1000}=2.5 \) kg.

Step3: Calculate temperature change

The initial temperature \( T_1 = 10.0^\circ\text{C} \), final temperature \( T_2 = 35.0^\circ\text{C} \). So \( \Delta T=T_2 - T_1=35.0 - 10.0 = 25.0^\circ\text{C} \).

Step4: Substitute values into the formula

We know \( Q = 143.75 \) kJ, \( m = 2.5 \) kg, \( \Delta T = 25.0^\circ\text{C} \). Plug into \( c=\frac{Q}{m\Delta T} \):
\( c=\frac{143.75}{2.5\times25.0} \)
First calculate the denominator: \( 2.5\times25.0 = 62.5 \)
Then \( c=\frac{143.75}{62.5}=2.3 \) kJ/kg\(^\circ\)C.

Answer:

The specific heat capacity of 1 - propanol is \( 2.3 \) kJ/kg\(^\circ\)C.