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:**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.**2. Recall the formula:** The formula relating heat (Q), mass (m), specific heat capacity (c), and change in temperature (ΔT) is:Q = mcΔT**3. Identify the knowns:*** Q (heat absorbed) = 7200 J* m (mass) = 300 g* ΔT (change in temperature) = 45°C - 15°C = 30°C**4. Solve for c (specific heat capacity):**Rearrange the formula to solve for c:c = Q / (mΔT)Substitute the known values:c = 7200 J / (300 g * 30°C)c = 7200 J / 9000 g°Cc = 0.8 J/g°C**Answer:** The specific heat capacity of the metal is 0.8 J/g°C.