Question

A large electromagnet draws 300A at 270V. The coils of the electromagnet are cooled by a flow of mystery liquid passing over them. The liquid enters the electromagnet at a temperature of 15 deg C, absorbs the Joule heat, and leaves at a higher temperature. If the liquid is to leave at a temperature no higher than 81 deg C, and the maximum flow rate of liquid through the electromagnet is 0.307kg/s. What is the specific heat of the mystery liquid? Pick the closest value. The solution should include the correct units.

Answer 1

Answer:
`c=3.99\approx 4\ kJ/kg-K`

Step-by-step explanation:

Given

Current
`I=300\ A`

Voltage
`V=270\ V`

Mass flow rate
`m=0.307\ kg/s`

Inlet temperature is
`T_i=15^(\circ)C\approx 288\ K`

`T_(out)\leq 81^(\circ)C`

Here heat of Electromagnet is absorbed by liquid

Heat rate of Electromagnet
`\dot{Q}=VI`

`\dot{Q}=270* 300=81\ kJ/s`

Heat absorbed by liquid
`\dot{Q}=mc(\Delta T)`

`\dot{Q}=0.307* c* (81-15)`

`81* 10^3=0.307* c* (81-15)`

`c=3.99\ kJ/kg-K`