Question

It takes \(880 \mathrm{J}\) to raise the temperature of \(350 \mathrm{g}\) of lead from 0 to \(20.0^{\circ} \mathrm{C} .\) What is the specific heat of lead?

Short answer

Answer: The specific heat of lead is \(0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\).

Step-by-step solution

Identify the given values

We are given the following information: - Amount of energy required to raise the temperature (Q) = 880 J - Mass of lead (m) = 350 g - Initial temperature (T1) = 0°C - Final temperature (T2) = 20.0°C

Write down the formula for energy change

The formula for energy change, which relates energy, mass, specific heat, and temperature change is: \(Q = mc (T_{2} - T_{1})\) We need to solve for the specific heat of lead (c), so we can rewrite the formula as follows: \(c = \frac{Q}{m(T_{2} - T_{1})}\)

Plug in the given values into the formula

Now we can substitute the given values into the equation: \(c = \frac{880 \mathrm{J}}{350 \mathrm{g}(20.0^{\circ}\mathrm{C}-0^{\circ}\mathrm{C})}\)

Calculate the specific heat

Simplify and compute the value for the specific heat: \(c = \frac{880 \mathrm{J}}{350 \mathrm{g}(20.0^{\circ}\mathrm{C})}\) \(c = \frac{880 \mathrm{J}}{7000 \mathrm{g}\cdot^{\circ}\mathrm{C}}\) \(c = 0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\)

Write the final answer

The specific heat of lead is \(0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\).