Certainly! Let's walk through the steps to calculate the heat absorbed by an aluminum alloy cylinder block when its temperature increases.
Given:
- Mass of the aluminum alloy cylinder block ([tex]\( m \)[/tex]) = 60 kg
- Initial temperature ([tex]\( T_i \)[/tex]) = 15°C
- Final temperature ([tex]\( T_f \)[/tex]) = 85°C
- Specific heat capacity ([tex]\( c \)[/tex]) = 0.92 kJ/kg·K
### Step 1: Determine the Temperature Change
First, we need to find the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[
\Delta T = T_f - T_i = 85^\circ C - 15^\circ C
\][/tex]
[tex]\[
\Delta T = 70^\circ C
\][/tex]
### Step 2: Calculate the Heat Absorbed
To find the amount of heat ([tex]\( Q \)[/tex]) absorbed by the cylinder block, we use the formula:
[tex]\[
Q = m \cdot c \cdot \Delta T
\][/tex]
where:
- [tex]\( Q \)[/tex] is the heat absorbed
- [tex]\( m \)[/tex] is the mass of the object
- [tex]\( c \)[/tex] is the specific heat capacity
- [tex]\( \Delta T \)[/tex] is the change in temperature
Plugging in the given values:
[tex]\[
Q = 60 \, \text{kg} \times 0.92 \, \text{kJ/kg·K} \times 70 \, K
\][/tex]
[tex]\[
Q = 60 \times 0.92 \times 70
\][/tex]
[tex]\[
Q = 3864 \, \text{kJ}
\][/tex]
So, the temperature change is [tex]\( 70^\circ C \)[/tex], and the heat absorbed by the aluminum alloy cylinder block is 3864 kJ.
