Solve Heating Water Problem: 100-W Electric Heater

[SpatialVacancy]

Hello all,

I do not know how to solve this proble and was wondering if you could offer some guidance.

Question:
How long after it is turned on will a 100-W electric
heater take to bring a quart of water to a boil from room
temperature 20 deg C?

Here's what I know:
$$T_i=20$$
$$T_f=100$$
$$\Delta T=80$$

$$100W = \frac{100J}{1s}$$

Thanks!

 

[recon]

I'm don't know what a quart is. However, I do know the specific heat capacity of water which is 4180J/kgK. This means that for every kilogram of water, 4180 joules of energy is needed to raise the temperature by 1 Kelvin or 1 degree celsius.

 

[BLaH!]

The key to this problem is knowing how much the temperature of water increases when a certain amount of heat is added. This is basically the definition of the specific heat:

$$c = \frac{1}{V}\frac{\Delta Q}{\Delta T} \Rightarrow \Delta Q = cV\Delta T$$

where $$c$$ is the specific heat and $$V$$ is the volume of the substance, in this case water. This equation is what you need to solve the problem. However, you are not told anything about the added heat. You are given information about the heater power. Assuming that all of the power goes into heating the water we can write:

$$P = \frac{\Delta Q}{\Delta t}=cV\frac{\Delta T}{\Delta t}$$

where $$\Delta t$$ is the time it takes to add the amount of heat $$\Delta Q$$. Just solve for $$\Delta t$$.

 

[Clausius2]

The key to this problem is knowing how much the temperature of water increases when a certain amount of heat is added. This is basically the definition of the specific heat:

$$c = \frac{1}{V}\frac{\Delta Q}{\Delta T} \Rightarrow \Delta Q = cV\Delta T$$

where $$c$$ is the specific heat and $$V$$ is the volume of the substance, in this case water. This equation is what you need to solve the problem. However, you are not told anything about the added heat. You are given information about the heater power. Assuming that all of the power goes into heating the water we can write:

$$P = \frac{\Delta Q}{\Delta t}=cV\frac{\Delta T}{\Delta t}$$

where $$\Delta t$$ is the time it takes to add the amount of heat $$\Delta Q$$. Just solve for $$\Delta t$$.

I think c has units of (J/m^3*K) for you. It is not the usual definition of the specific heat which is always written in term of masses (J/KgK). Anyway you have solved very good the problem indeed, in my opinion.

If you want and advice never write again $$\Delta Q$$. The "increment" or "variation" of heat has none physical meaning. It has no sense. The heat does not vary. A body does not have any heat itself. So that no variation is possible. Instead of it, write simply $$Q$$ or $$\delta Q$$ (inexact differential) when you talk about heat.

It was only a some clearings, in my opinion your solution sounds very good.