Question

Merlinda is heating 2-kg water initially at 6.5 °C using a 100-Watt solar-powered heater. How long will it take (in seconds) to reach the boiling point?

Answer 1

Calculating the Kilowatt-hours(kWh) required to heat the water using the following formula:

$Pt = (4.2 \times L \times \Delta T) \div 3600 \\ where: \\ Pt = the \ power \ used \ to \ heat \ the \ water, in \ kWh, \\ L = number \ of \ liters \ of \ water \\ \Delta T = difference \ in \ temperature \ (in \ ^\circ C) \\ mass = 2kg = 2liters \\ Pt = (4.2 \times 2 \times (100 - 6.5)) \div 3600 \\ Pt = 0.218 \ kWh \\$

To calculate the amount of time it takes to heat the water, we divide the power used to heat the water (0.218) by the heater element rating, listed in kW.

$$Heating \ time = \frac{Power \ used \ to \ heat \ the \ water, \ in \ kWh}{Heat \ element \ rating, \ in \ kW}$$

Power used to heat the water = 0.218 kWh

Heat element rating = 100 W = 100/1000 kW = 0.1 kW

$\displaystyle Heating \ time = \frac{0.218 kWh}{0.1 kW} = 2.18 \ hrs \\ Heating \ time = 7848 \ s \ (in \ seconds)$

OR, SOLVING USING THE FORMULA OF HEAT CAPACITY:

$\displaystyle \Delta E = mc \Delta\theta \\ \Delta E = mc(\theta_2 - \theta_1) \\ where, \Delta E = Thermal \ energy \\ m = mass \ of \ the \ water = 2kg = 2litres\\ c = specific \ heat \ capacity \ of \ water \\ \Delta\theta = change \ in \ temperature \\ \\ c = 4186J/kg/K \\ \\ \Delta E = mc(\theta_2 - \theta_1) \\ \Delta E = 2 \times 4186 \times (100^\circ C - 6.5^\circ C) \\ \Delta E = 782782J \\ Power, P = \frac {Change \ in \ Thermal \ Energy, \Delta E}{Time, t} \\ P = \frac {782782 J}{3600s} \\ P = 217.44 \ W \\ \therefore \\ Time \ required \ with \ 100W \ heater = \frac{217.44}{100} \\ t = 2.1744 \ hrs = 7827.84\ seconds \\ t = 7828 \ s$