Question

A small immersion heater is rated at 345 W. The specfic heat of water is 4186 J/kg x C degrees.

<Ch 14 HW Item 5 5 or 10 Constants A small immersion heater is rated at 345 W. The specific heat of water is 4186J/kg-C Part A Estimate how long it will take to heat a cup of soup (assume this is 250 mL of water from 15 to 63 C.Ignore the heat loss to the surrounding environment. Express your answer using two significant figures. Η ΑΣΦ 6. ? Submit Previous Answers Request Answer X Incorrect; Try Again Provide Feedback Next >

Answer 1

P = power of heating heater = 345 W Pa olt Q = heat transfer MCAT me mass of soup = V XS V volume of soup (water) = 2some } = density of water VXS- = 0.25 kg c = specific beat of water = 4186 Jlig į AT - change in temperature (63-15) = 0.001 Kg / me m 250 X 0.001 - = 48°C 0 0.25 X 4186 X 48 :. t = m ĊAT Р A 345 ta 1.5 x 18s This is the time to take to heat the soup from temperature 15°c to 63°c

Answer 2

SOLUTION :

Mass of 250 mL water = 250 g = 0.25 kg

Heat required to heat this water to 63ºC from 15ºC

= m s (t2 - t1)

= 0.25 * 4186 * (63 - 15) 

= 50232 J 

Let time taken is t sec by the 345 W immersion heater .

Electrical energy consumed

 = 345 * t      watt-sec 

= 345 t        J ( since 1 watt-sec = 1 J)

As per conservation of energy :

Electrical energy consumed = Heat energy generated to heat the water

=> 345 t = 50232 

=> t = 50232 / 345 = 145.60 sec = 2 min. 25,60 secs.

Time needed to heat the water as required = 145.60 secs = 2 minutes 25.60 secs (ANSWER).