14. A 30 g block of a metal absorbs 7200 J of heat as its temperature increases from 15^circ C to 45^circ C What is th specific heat capacity of the metal?

Here's how to calculate the specific heat capacity of the metal:<br /><br />**1. Understand the concept:** Specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.<br /><br />**2. Recall the formula:** The formula relating heat (Q), mass (m), specific heat capacity (c), and change in temperature (ΔT) is:<br /><br />Q = mcΔT<br /><br />**3. Identify the knowns:**<br /><br />* Q (heat absorbed) = 7200 J<br />* m (mass) = 300 g<br />* ΔT (change in temperature) = 45°C - 15°C = 30°C<br /><br />**4. Solve for c (specific heat capacity):**<br /><br />Rearrange the formula to solve for c:<br /><br />c = Q / (mΔT)<br /><br />Substitute the known values:<br /><br />c = 7200 J / (300 g * 30°C)<br /><br />c = 7200 J / 9000 g°C<br /><br />c = 0.8 J/g°C<br /><br />**Answer:** The specific heat capacity of the metal is 0.8 J/g°C.<br />