Answer :
To solve this problem, we need to determine the specific heat capacity of the object. The specific heat capacity can be found using the formula:
[tex]\[
c = \frac{Q}{m \cdot \Delta T}
\][/tex]
Where:
- [tex]\( c \)[/tex] is the specific heat capacity.
- [tex]\( Q \)[/tex] is the amount of heat energy absorbed or released, in joules (J).
- [tex]\( m \)[/tex] is the mass of the object, in kilograms (kg).
- [tex]\( \Delta T \)[/tex] is the change in temperature, in degrees Celsius ([tex]\(^{\circ}C\)[/tex]).
Given the values:
- [tex]\( Q = 67500 \, \text{J} \)[/tex]
- [tex]\( m = 5.0 \, \text{kg} \)[/tex]
- [tex]\( \Delta T = 30 \, ^{\circ}C \)[/tex]
We can substitute these values into the formula to find the specific heat capacity:
[tex]\[
c = \frac{67500}{5.0 \times 30}
\][/tex]
When you perform the calculation:
1. Multiply the mass [tex]\( m \)[/tex] by the temperature change [tex]\( \Delta T \)[/tex]:
[tex]\[ 5.0 \times 30 = 150 \][/tex]
2. Divide the heat energy [tex]\( Q \)[/tex] by the result from step 1:
[tex]\[ c = \frac{67500}{150} \][/tex]
3. Calculate the result from step 2:
[tex]\[ c = 450.0 \, \text{J/kg}^{\circ}C \][/tex]
Therefore, the specific heat capacity of the object is [tex]\( 450.0 \, \text{J/kg}^{\circ}C \)[/tex].
[tex]\[
c = \frac{Q}{m \cdot \Delta T}
\][/tex]
Where:
- [tex]\( c \)[/tex] is the specific heat capacity.
- [tex]\( Q \)[/tex] is the amount of heat energy absorbed or released, in joules (J).
- [tex]\( m \)[/tex] is the mass of the object, in kilograms (kg).
- [tex]\( \Delta T \)[/tex] is the change in temperature, in degrees Celsius ([tex]\(^{\circ}C\)[/tex]).
Given the values:
- [tex]\( Q = 67500 \, \text{J} \)[/tex]
- [tex]\( m = 5.0 \, \text{kg} \)[/tex]
- [tex]\( \Delta T = 30 \, ^{\circ}C \)[/tex]
We can substitute these values into the formula to find the specific heat capacity:
[tex]\[
c = \frac{67500}{5.0 \times 30}
\][/tex]
When you perform the calculation:
1. Multiply the mass [tex]\( m \)[/tex] by the temperature change [tex]\( \Delta T \)[/tex]:
[tex]\[ 5.0 \times 30 = 150 \][/tex]
2. Divide the heat energy [tex]\( Q \)[/tex] by the result from step 1:
[tex]\[ c = \frac{67500}{150} \][/tex]
3. Calculate the result from step 2:
[tex]\[ c = 450.0 \, \text{J/kg}^{\circ}C \][/tex]
Therefore, the specific heat capacity of the object is [tex]\( 450.0 \, \text{J/kg}^{\circ}C \)[/tex].
