Question

An electrical heater is used to heat 100g of water in a well-insulated container at a steady rate. The temperature of the water increases by 15 degrees C when the heater is operated for a period of 5.0 minutes. Determine the change of temperature of the water when the same heater and container are individually used to heat (a) 300g of water in 5 mins and (b)100g of water for a time of 2.5 minutes.

Answer 1

you need to find the power of this heater.
\[Q=mc \theta \]
\[Q=Pt\]
Q=heat energy, P=power, t=time, m=mass, t=time taken, delta=temperature change, c=specific heat capacity of water

so,
\[Pt=mc\theta\]
\[P=(mc \theta )/t\]
Now you can find the power of this heater using the formula given. After finding the power, you can use these equations to get your answer for question a) and b).

Answer 2

What does θ represent ?

Answer 3

change in temperature

Answer 4

Ok thanks

Answer 5

3780000 J ?

Answer 6

So what next. Can i use ΔT = Q / mc.

Answer 7

how you got 378000J?
water specific heat capacity is 4.18 J g^-1 C^-1
P=(100g *4.18 J g^-1 C^-1* 15 C)/ ( 5*60s) =20.9 W

yes you can use ΔT = Q / mc. But for me, I will use ΔT = Pt / mc because i have different heating time so my heat energy is different in each cases.

Answer 8

yup

Answer 9

To calculate the power of the heater P =( mcθ) / t - Mass = 0.1kg* The correct specific heat capacity for water 4200 J kg-1 K-1 * 15°C / 300seconds = 21 W for the power of the heater.

Answer 10

So what formula do i now use to calculate the change in temperature?

Answer 11

ΔT = Pt / mc

Answer 12

a) = 8.8 degrees C Temperature change
b ) = 3.2 degrees C temperature change ?

Answer 13

a) ΔT = Pt / mc=(21 W* 300s)/(300g *4.2 J g^-1 C^-1) =5 degreee celecius
b) ΔT = Pt / mc = (21W *150 s)/ (100g *4.2 J g^-1 C^-1) =7.5 degree celcius