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 300g of water for the same period of time: Help!!

Answer 1

I don't get the "individually" in the wording of the question!

Answer 2

It probably means that the same container is used again?

Answer 3

Compute the power used first.

Answer 4

Ok. So that is ∆Q/t ?

Answer 5

But i can't calculate the power when i don't have the energy which supplies the heater!

Answer 6

you know the energy used. \[\Delta Q = mc \Delta T \]

Answer 7

Ok. So Q=0.3kg*4200Jkg^-1k^-1* ?

Answer 8

um, yes.. actually you don't need to know the power. I read it again, and it's for the same amount of time, so it's the same energy. i.e. Q is the same, so you just have a different Delta T for the different mass. You can solve it with a proportion.

Answer 9

\[\Delta T _{2}=m_1T_1/m_2\]

Answer 10

Or, more explicitly - \[\Delta Q_1=\Delta Q_2\]
\[\Delta Q=mc \Delta T\]
\[m_1 c \Delta T_1=m_2 c \Delta T_2 \rightarrow m_1 \Delta T_1=m_2 \Delta T_2\]

Answer 11

I was thinking this:∆Q =MwCw∆ᶱ?

Answer 12

Is that a delta-theta?
The form is right though: that is the governing equation, but since Q and c are the same in both situations, you only need to consider the ratios of mass and temperature.

Answer 13

What are T1 and T2?

Answer 14

T1= temperature one (given); T2= temperature two (that you are solving for)

Answer 15

Help!

Answer 16

?

Answer 17

What more do you need help with? You have the answer.

Answer 18

I calculate a 5 degree celsius change of temperature of the water. Is that correct

Answer 19

yes. three times the mass yields one-third the amount of temperature change since it is an inverse relation.

Answer 20

Ok thanks. What if the water was 100g and the time 2.5 minutes and i had to calculate the change in temperature?

Answer 21

If the new time was 2.5min?
Then with same power input, half the time gives half the energy, so divide Q1 by 2.