the specific heat of a certain type of metal is 0.128 J(grams per Celsius). What is the final temperature if 305J of heat is added to 74.8g of this metal initially at 20.0 degrees Celsius?
q = mass x specific heat x (Tfinal-Tinitial)
45.8 Degrees C
To find the final temperature, we can use the equation:
\(q = mcΔT\)
Where:
- \(q\) is the heat energy absorbed or released by the substance
- \(m\) is the mass of the substance
- \(c\) is the specific heat capacity of the substance
- \(\Delta T\) is the change in temperature
We can rearrange the equation to solve for the change in temperature:
\(\Delta T = \frac{q}{mc}\)
Given:
- \(q = 305 \, \text{J}\)
- \(m = 74.8 \, \text{g}\)
- \(c = 0.128 \, \text{J/(g°C)}\)
Substituting these values into the equation:
\(\Delta T = \frac{305 \, \text{J}}{74.8 \, \text{g} \times 0.128 \, \text{J/(g°C)}}\)
Simplifying the calculation:
\(\Delta T = \frac{305}{9.5744}\)
\(\Delta T \approx 31.8594 \, \text{°C}\)
Now, to find the final temperature, we add the change in temperature to the initial temperature:
\(T_{\text{final}} = T_{\text{initial}} + \Delta T\)
\(T_{\text{final}} = 20.0°C + 31.8594°C\)
\(T_{\text{final}} \approx 51.86°C\)
Therefore, the final temperature is approximately 51.86°C.