Answer :
To determine the specific heat ([tex]\(C_p\)[/tex]) of the substance, we will use the formula for calculating the heat applied:
[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]
where:
- [tex]\( q \)[/tex] represents the heat applied,
- [tex]\( m \)[/tex] is the mass of the substance,
- [tex]\( \Delta T \)[/tex] is the change in temperature,
- [tex]\( C_p \)[/tex] is the specific heat.
Given values:
- [tex]\( q = 2520 \, \text{J} \)[/tex]
- [tex]\( m = 10.0 \, \text{kg} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 10.0^{\circ} \text{C} \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 70.0^{\circ} \text{C} \)[/tex]
First, calculate the change in temperature, [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]
[tex]\[ \Delta T = 70.0^{\circ} \text{C} - 10.0^{\circ} \text{C} \][/tex]
[tex]\[ \Delta T = 60.0^{\circ} \text{C} \][/tex]
Next, we substitute the given values into the formula and solve for [tex]\( C_p \)[/tex]:
[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]
[tex]\[ 2520 \, \text{J} = 10.0 \, \text{kg} \cdot C_p \cdot 60.0^{\circ} \text{C} \][/tex]
Rearrange to solve for [tex]\( C_p \)[/tex]:
[tex]\[ C_p = \frac{2520 \, \text{J}}{10.0 \, \text{kg} \cdot 60.0^{\circ} \text{C}} \][/tex]
[tex]\[ C_p = \frac{2520 \, \text{J}}{600 \, \text{kg} \cdot ^{\circ} \text{C}} \][/tex]
[tex]\[ C_p = 4.2 \, \text{J/(kg} \cdot ^{\circ} \text{C)} \][/tex]
Since the choices provided are in [tex]\( \text{J/(g} \cdot ^{\circ} \text{C)} \)[/tex], we need to convert [tex]\( C_p \)[/tex] from [tex]\( \text{J/(kg} \cdot ^{\circ} \text{C)} \)[/tex] to [tex]\( \text{J/(g} \cdot ^{\circ} \text{C)} \)[/tex].
We know that:
[tex]\[ 1 \, \text{kg} = 1000 \, \text{g} \][/tex]
So,
[tex]\[ C_p = \frac{4.2 \, \text{J/(kg} \cdot ^{\circ} \text{C)}}{1000} \][/tex]
[tex]\[ C_p = 0.0042 \, \text{J/(g} \cdot ^{\circ} \text{C)} \][/tex]
Thus, the specific heat of the substance is:
[tex]\[ \boxed{0.00420 \, \text{J/(g} \cdot ^{\circ} \text{C)}} \][/tex]
[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]
where:
- [tex]\( q \)[/tex] represents the heat applied,
- [tex]\( m \)[/tex] is the mass of the substance,
- [tex]\( \Delta T \)[/tex] is the change in temperature,
- [tex]\( C_p \)[/tex] is the specific heat.
Given values:
- [tex]\( q = 2520 \, \text{J} \)[/tex]
- [tex]\( m = 10.0 \, \text{kg} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 10.0^{\circ} \text{C} \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 70.0^{\circ} \text{C} \)[/tex]
First, calculate the change in temperature, [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]
[tex]\[ \Delta T = 70.0^{\circ} \text{C} - 10.0^{\circ} \text{C} \][/tex]
[tex]\[ \Delta T = 60.0^{\circ} \text{C} \][/tex]
Next, we substitute the given values into the formula and solve for [tex]\( C_p \)[/tex]:
[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]
[tex]\[ 2520 \, \text{J} = 10.0 \, \text{kg} \cdot C_p \cdot 60.0^{\circ} \text{C} \][/tex]
Rearrange to solve for [tex]\( C_p \)[/tex]:
[tex]\[ C_p = \frac{2520 \, \text{J}}{10.0 \, \text{kg} \cdot 60.0^{\circ} \text{C}} \][/tex]
[tex]\[ C_p = \frac{2520 \, \text{J}}{600 \, \text{kg} \cdot ^{\circ} \text{C}} \][/tex]
[tex]\[ C_p = 4.2 \, \text{J/(kg} \cdot ^{\circ} \text{C)} \][/tex]
Since the choices provided are in [tex]\( \text{J/(g} \cdot ^{\circ} \text{C)} \)[/tex], we need to convert [tex]\( C_p \)[/tex] from [tex]\( \text{J/(kg} \cdot ^{\circ} \text{C)} \)[/tex] to [tex]\( \text{J/(g} \cdot ^{\circ} \text{C)} \)[/tex].
We know that:
[tex]\[ 1 \, \text{kg} = 1000 \, \text{g} \][/tex]
So,
[tex]\[ C_p = \frac{4.2 \, \text{J/(kg} \cdot ^{\circ} \text{C)}}{1000} \][/tex]
[tex]\[ C_p = 0.0042 \, \text{J/(g} \cdot ^{\circ} \text{C)} \][/tex]
Thus, the specific heat of the substance is:
[tex]\[ \boxed{0.00420 \, \text{J/(g} \cdot ^{\circ} \text{C)}} \][/tex]
