Question

An automatic coffee maker uses a resistive heating element to boil the 2.4 kg of water that was poured into it at 21 °C. The current delivered to the coffee pot is 8.5 A when it is plugged into a 120 V electrical outlet. If the specific heat capacity of water is 4186 J/kgC° , approximately how long does it take to boil all of the water?

Answer 1

Answer:

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,  

$\Delta H$

is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$

is the temperature change

Thus, given that:-

Mass of water = 2.4 kg

Specific heat = 4.18 J/g°C

$\Delta T=100-21\ ^0C=79\ ^0C$

So,  

$\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ$

Heat Supplied $Q=VIt$

where $i=current$

$V=Voltage$

$8.5\times 120\times t=2.4\times 4186\times 79$

$t=778.10 s$