Question

A 200.0 - \(\mu\) F capacitor is placed across a \(12.0-\mathrm{V}\) battery. When a switch is thrown, the battery is removed from the capacitor and the capacitor is connected across a heater that is immersed in $1.00 \mathrm{cm}^{3}$ of water. Assuming that all the energy from the capacitor is delivered to the water, what is the temperature change of the water?

Short answer

Answer: The temperature change of the water is approximately 3.44°C.

Step-by-step solution

Calculate the initial energy stored in the capacitor

To calculate the initial energy stored in the capacitor, we can use the formula for the energy stored in a capacitor: \(E = \frac{1}{2} C V^2\) Where E is the energy, C is the capacitance, and V is the voltage. In this problem, we are given C = 200.0 µF and V = 12.0 V. Let's calculate E. Solution: \(E = \frac{1}{2} (200.0 \times 10^{-6}\,\text{F}) (12.0\,\text{V})^2\) \(E = 0.5 \times 200.0 \times 10^{-6}\,\text{F} \times 144\,\text{V}^2\) \(E \approx 14.4\,\text{mJ}\)

Calculate the energy required to heat the water

Next, we'll calculate the energy required to heat the water using the specific heat capacity formula: \(E = mc\Delta T\) Where E is the energy, m is the mass, c is the specific heat capacity, and \(\Delta T\) is the temperature change. In this problem, we have a volume of 1.00 cm³ of water, and to convert volume to mass, we need to multiply by the density of water, which is 1 g/cm³. Therefore, m = 1.00 g. The specific heat capacity of water is 4.186 J/g°C. Since we know the energy of the capacitor calculated in step 1, we can rewrite the formula like this: \(\Delta T = \frac{E}{mc}\)

Relate the energy from the capacitor to the energy required to heat the water and calculate the temperature change

Now we'll input the values for E, m and c into the formula and calculate the temperature change \(\Delta T\). \(\Delta T = \frac{14.4 \times 10^{-3}\, \text{J}}{(1.00\,\text{g})(4.186\,\text{J/g°C})}\) \(\Delta T \approx 3.44\,\text{°C}\) The temperature change of the water is approximately 3.44°C.