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

Step-by-step explanation:

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

`\Delta H=m* C* \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* 4.18* 79\ J=792.52\ kJ`

Heat Supplied
`Q=VIt`

where
`i=current`

`V=Voltage`

`8.5* 120* t=2.4* 4186* 79`

`t=778.10 s`