Energy basic question 2

[Dhamnekar Winod]

Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F.
Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F (29.44 °C) to 72 °F (22.22 °C) in 24 hours. Estimate
the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that \(1
kilowatt-hour (kWh) = 3.6 \times 10^6 J,\) and assume that the density of water is 1.0 g/mL (independent of
temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were
they likely to yield “positive” or “negative” errors)?

Answer: To estimate the amount of electrical energy required to keep the waterbed from cooling, we need to calculate the heat loss and then convert it to kilowatt-hours (kWh).

1. **Calculate the mass of the water**:
- Volume of water = 892 L = 892,000 mL
- Density of water = 1.0 g/mL
- Mass of water = 892,000 g = 892 kg

2. **Calculate the temperature change**:
- Initial temperature = 85 °F
- Final temperature = 72 °F
- Temperature change (ΔT) in Celsius:
\[
\Delta T = \frac{85 - 32}{1.8} - \frac{72 - 32}{1.8} = 29.44 - 22.22 = 7.22 \, \text{°C}
\]

3. **Calculate the heat loss**:
- Specific heat capacity of water (c) = 4.18 J/g°C
- Heat loss (Q) = mass × specific heat capacity × temperature change
\[
Q = 892,000 \, \text{g} \times 4.18 \, \text{J/g°C} \times 7.22 \, \text{°C} = 26,936,656 \, \text{J}
\]

4. **Convert the heat loss to kilowatt-hours**:
- 1 kWh = 3.6 × 10^6 J
\[
\text{Energy required (kWh)} = \frac{26,936,656 \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} \approx 7.48 \, \text{kWh}
\]

### Assumptions and Their Effects:
1. Density of water is constant: Assuming the density of water is 1.0 g/mL regardless of temperature simplifies the calculation. In reality, density slightly changes with temperature, but this assumption likely has a negligible effect on the result.
2. Specific heat capacity is constant: Assuming the specific heat capacity of water is constant at 4.18 J/g°C simplifies the calculation. The specific heat capacity can vary slightly with temperature, but this assumption is reasonable and introduces minimal error.
3. No heat loss to the environment: Assuming all the heat loss is only due to the cooling of water and not to the environment. This assumption might lead to a slight underestimation of the energy required, as some heat would be lost to the surroundings.

These assumptions are generally reasonable and introduce minor errors, making the calculated energy requirement a good estimate.