Answer:
The temperature change is 96.97°C.
Step-by-step explanation:
The given information from the exercise is:
- Mass (m): 150g
- Specific heat of mercury (c): 0.0330cal/g.°C
- Heat (q): 480cal
To calculate the temperature change (ΔT) it is necessary to use the Heat formula, and replace the values of mass, specific heat and heat:
![\begin{gathered} q=m*c*\Delta T \\ 480cal=150g*0.0330(cal)/(g*°C)*\Delta T \\ 480cal=4.95(cal)/(°C)*\Delta T \\ (480cal)/(4.95(cal)/(°C))=\Delta T \\ 96.97\degree C=\Delta T \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/tb6aimeolfmv93y4scxx16xvhtn06mj3qz.png)
So, the temperature change is 96.97°C.