We are tasked with finding the final temperature of the metal block. Here's the breakdown of the given data:

• Mass of the block: $m = 25.0 \, \text{g}$
• Energy absorbed: $q = 255 \, \text{cal}$ (convert to joules: $1 \, \text{cal} = 4.184 \, \text{J}$)
• Initial temperature: $T_{\text{initial}} = 17.0^\circ \, \text{C}$
• Specific heat capacity: $c = 2.72 \, \text{J/g}^\circ \, \text{C}$

We use the formula for heat transfer: $q = m \cdot c \cdot \Delta T$ Where: $\Delta T = T_{\text{final}} - T_{\text{initial}}$

First, convert $q$ to joules: $q = 255 \, \text{cal} \times 4.184 \, \text{J/cal} = 1067.92 \, \text{J}$

Now solve for $T_{\text{final}}$: $1067.92 = 25.0 \cdot 2.72 \cdot (T_{\text{final}} - 17.0)$

Simplify: $1067.92 = 68.0 \cdot (T_{\text{final}} - 17.0)$

Divide through by 68.0: $\frac{1067.92}{68.0} = T_{\text{final}} - 17.0$ $15.7 = T_{\text{final}} - 17.0$

Solve for $T_{\text{final}}$: $T_{\text{final}} = 15.7 + 17.0 = 32.7^\circ \, \text{C}$

Final Answer:

The final temperature of the metal block is 32.7°C.

Let me know if you'd like more details or have additional questions! Here are some related questions to explore:

1. How is the specific heat capacity related to the thermal properties of a material?
2. What would happen if the mass of the block was doubled in this problem?
3. How does converting calories to joules affect calculations?
4. What are practical applications of this specific heat formula in everyday scenarios?
5. Can this method be used to calculate temperature changes in liquids or gases?

Tip: Always make sure to convert units properly in thermal physics problems to avoid calculation errors!