Questions: A 55.0 kilogram (kg) block of metal has an original temperature of 15.0°C and 0.45 J/g°C. What will be the final temperature of this metal if 75500 J of heat energy is absorbed by the metal?

Solution Steps

We are given:

• Mass of the metal block, \( m = 55.0 \, \text{kg} \)
• Initial temperature, \( T_i = 15.0^{\circ} \text{C} \)
• Specific heat capacity, \( c = 0.45 \, \frac{\text{J}}{\text{g}^{\circ} \text{C}} \)
• Heat energy absorbed, \( Q = 75500 \, \text{J} \)

Since the specific heat capacity is given in \(\frac{\text{J}}{\text{g}^{\circ} \text{C}}\), we need to convert the mass from kilograms to grams: \[ m = 55.0 \, \text{kg} \times 1000 \, \frac{\text{g}}{\text{kg}} = 55000 \, \text{g} \]

The formula to calculate the change in temperature when heat energy is absorbed is: \[ Q = mc\Delta T \] where \(\Delta T\) is the change in temperature.

Rearrange the formula to solve for \(\Delta T\): \[ \Delta T = \frac{Q}{mc} \]

Substitute the given values: \[ \Delta T = \frac{75500 \, \text{J}}{55000 \, \text{g} \times 0.45 \, \frac{\text{J}}{\text{g}^{\circ} \text{C}}} \]

\[ \Delta T = \frac{75500}{55000 \times 0.45} \] \[ \Delta T = \frac{75500}{24750} \] \[ \Delta T \approx 3.051 \, ^{\circ} \text{C} \]

The final temperature \( T_f \) is the initial temperature plus the change in temperature: \[ T_f = T_i + \Delta T \] \[ T_f = 15.0^{\circ} \text{C} + 3.051^{\circ} \text{C} \] \[ T_f \approx 18.051^{\circ} \text{C} \]

Final Answer