Question

(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is 9.00¢/(kW · h) and the specific heat for water is 4184 J/(kg · °C). $ 67 Incorrect: Your answer is incorrect. How much heat is needed to raise the temperature of m kg of a substance? How many joules are in 1 kWh? (b) What current was used by the 220 V AC electric heater, if this took 3.45 h? 88.2 Correct: Your answer is correct. A

Answer 1

Answer:

a) $E = 6.024\,USD$, For m kilograms, it is 4184m J., 3600000 joules, b) $i = 88.200\,A$

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

$Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})$

Where:

$m_{w}$ - Mass of water, measured in kilograms.

$c_{w}$ - Specific heat of water, measured in $\frac{J}{kg\cdot ^{\circ}C}$.

$T_{f}$, $T_{i}$ - Initial and final temperatures, measured in $^{\circ}C$.

Then,

$Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)$

$Q_{needed} = 180748800\,J$

The energy needed in kilowatt-hours is:

$Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)$

$Q_{needed} = 50.208\,kWh$

The electric energy required to heat up the water is:

$E = \frac{50.208\,kWh}{0.75}$

$E = 66.944\,kWh$

Lastly, the cost of heating a hot tub is: (USD - US dollars)

$E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)$

$E = 6.024\,USD$

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

$i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}$

$i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}$

$i = 88.200\,A$