A 5.25 g sample of metal gives off 10.4 J of energy as it cools from [tex]$49.5^{\circ} C$[/tex] to [tex]$40.5^{\circ} C$[/tex]. What is the specific heat of the metal?

[tex]\[ c = \left[? \frac{J}{g \cdot^{\circ} C}\right] \][/tex]

Note: [tex]q = -10.4 J[/tex]



Answer :

To find the specific heat of the metal, we can use the formula for heat transfer:

[tex]\[ q = mc\Delta T \][/tex]

where:
- [tex]\( q \)[/tex] is the heat energy transferred (in Joules),
- [tex]\( m \)[/tex] is the mass of the substance (in grams),
- [tex]\( c \)[/tex] is the specific heat capacity (in [tex]\( \frac{J}{g \cdot ^\circ C} \)[/tex]),
- [tex]\( \Delta T \)[/tex] is the change in temperature (in [tex]\( ^\circ C \)[/tex]).

Let's break down the problem step-by-step to find the specific heat capacity [tex]\( c \)[/tex]:

1. Identify the given values:
- The energy released: [tex]\( q = 10.4 \, \text{J} \)[/tex]
- The mass of the sample: [tex]\( m = 5.25 \, \text{g} \)[/tex]
- The initial temperature: [tex]\( T_{\text{initial}} = 49.5^\circ \text{C} \)[/tex]
- The final temperature: [tex]\( T_{\text{final}} = 40.5^\circ \text{C} \)[/tex]

2. Calculate the change in temperature [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = T_{\text{initial}} - T_{\text{final}} \][/tex]
[tex]\[ \Delta T = 49.5^\circ \text{C} - 40.5^\circ \text{C} = 9.0^\circ \text{C} \][/tex]

3. Rearrange the Heat Transfer Formula to solve for specific heat [tex]\( c \)[/tex]:
[tex]\[ c = \frac{q}{m \Delta T} \][/tex]

4. Substitute the known values into the equation:
[tex]\[ c = \frac{10.4 \, \text{J}}{5.25 \, \text{g} \cdot 9.0^\circ \text{C}} \][/tex]

5. Perform the calculation:
[tex]\[ c = \frac{10.4}{5.25 \times 9.0} \][/tex]
[tex]\[ c \approx \frac{10.4}{47.25} \approx 0.2201 \, \frac{J}{g \cdot ^\circ C} \][/tex]

Therefore, the specific heat capacity of the metal is approximately [tex]\( 0.2201 \frac{J}{g \cdot ^\circ C} \)[/tex].