Answer :
To find the final temperature of the metal after adding the heat, we can apply the formula related to heat transfer. Let's break it down step by step:
1. Understand the formula:
[tex]\[
Q = m \times c \times \Delta T
\][/tex]
where [tex]\( Q \)[/tex] is the heat added (in Joules), [tex]\( m \)[/tex] is the mass (in grams), [tex]\( c \)[/tex] is the specific heat capacity (in J/g°C), and [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
2. Rearrange the formula to solve for [tex]\(\Delta T\)[/tex]:
[tex]\[
\Delta T = \frac{Q}{m \times c}
\][/tex]
Here, [tex]\(\Delta T\)[/tex] represents the difference between the final temperature ([tex]\(T_{\text{final}}\)[/tex]) and the initial temperature ([tex]\(T_{\text{initial}}\)[/tex]).
3. Plug in the given values:
- Heat added, [tex]\( Q = 305 \, \text{J} \)[/tex]
- Mass, [tex]\( m = 39.3 \, \text{g} \)[/tex]
- Specific heat capacity, [tex]\( c = 0.128 \, \text{J/g°C} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 20.0 \, ^\circ \text{C} \)[/tex]
4. Calculate [tex]\(\Delta T\)[/tex]:
[tex]\[
\Delta T = \frac{305}{39.3 \times 0.128}
\][/tex]
[tex]\[
\Delta T \approx 60.63 \, ^\circ \text{C}
\][/tex]
5. Find the final temperature:
[tex]\[
T_{\text{final}} = T_{\text{initial}} + \Delta T
\][/tex]
[tex]\[
T_{\text{final}} = 20.0 \, ^\circ \text{C} + 60.63 \, ^\circ \text{C}
\][/tex]
[tex]\[
T_{\text{final}} \approx 80.63 \, ^\circ \text{C}
\][/tex]
So, the final temperature of the metal after adding 305 J of heat is approximately [tex]\(80.63 \, ^\circ \text{C}\)[/tex].
1. Understand the formula:
[tex]\[
Q = m \times c \times \Delta T
\][/tex]
where [tex]\( Q \)[/tex] is the heat added (in Joules), [tex]\( m \)[/tex] is the mass (in grams), [tex]\( c \)[/tex] is the specific heat capacity (in J/g°C), and [tex]\( \Delta T \)[/tex] is the change in temperature (in °C).
2. Rearrange the formula to solve for [tex]\(\Delta T\)[/tex]:
[tex]\[
\Delta T = \frac{Q}{m \times c}
\][/tex]
Here, [tex]\(\Delta T\)[/tex] represents the difference between the final temperature ([tex]\(T_{\text{final}}\)[/tex]) and the initial temperature ([tex]\(T_{\text{initial}}\)[/tex]).
3. Plug in the given values:
- Heat added, [tex]\( Q = 305 \, \text{J} \)[/tex]
- Mass, [tex]\( m = 39.3 \, \text{g} \)[/tex]
- Specific heat capacity, [tex]\( c = 0.128 \, \text{J/g°C} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 20.0 \, ^\circ \text{C} \)[/tex]
4. Calculate [tex]\(\Delta T\)[/tex]:
[tex]\[
\Delta T = \frac{305}{39.3 \times 0.128}
\][/tex]
[tex]\[
\Delta T \approx 60.63 \, ^\circ \text{C}
\][/tex]
5. Find the final temperature:
[tex]\[
T_{\text{final}} = T_{\text{initial}} + \Delta T
\][/tex]
[tex]\[
T_{\text{final}} = 20.0 \, ^\circ \text{C} + 60.63 \, ^\circ \text{C}
\][/tex]
[tex]\[
T_{\text{final}} \approx 80.63 \, ^\circ \text{C}
\][/tex]
So, the final temperature of the metal after adding 305 J of heat is approximately [tex]\(80.63 \, ^\circ \text{C}\)[/tex].