hi i have sent back this question because i have not undertood where does it come 3.91 Kw? so that was 3.1KW.

could please explain it ?

If a **3.1 kW **heater is used to heat up **0.5 kg** of water, how long will it take for the temperature to increase by **10℃**. Express answer in second with 2 decimal places.

Given: the specific heat of water is **4200 J / kg K**. and heat loss is neglected.

### 1 Answer | Add Yours

Now we know that 1 W is 1 J per second. 1kW = 1000 J / s.

The capacity of the heater is 3.1 kW or 3.1 kJ / s.

We need to increase the tempereature of 0.5 kg of water by 10 degree Celsius. The specific heat of water is given as 4200 J / kg*K.

That means 4200 J or 4.2 kJ are required to increase the temperature of 1 kg of water by 1 K.

We have to increase the temperature of 0.5 kg of water by 10 degree Celsius, that is the same as increasing the temperature by 10 K.

Therefore to do that, we need 4.2*0.5*10 kJ of energy.

So the energy required is 21 kJ.

Now the heater produces 3.1 kJ of energy every second. To produce 21 kJ of energy the time required is 21/3.1= 6.774 s

**Therefore the answer is 6.77 s.**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes