# How many heaters needed

A room 10"x10"x17" lxbxh respectively to be heated 397F if a heater is placed inside the room (2 kw) how many heater are to be required to heat the room from room temperature to 397 F in 1 hour.

• A ten inch by ten inch by seventeen inch room will not be able to contain much of a heater. You will also have to include the rate of energy loss this room experiences in order to have a suitable answer. Sep 4 '19 at 11:09
• Can you please provide what work you have already tried to solve this problem? As it is now, anyone who answers would be doing all of the work for you. Please help us to understand what parts of the task you already understand and what parts you are having trouble with.
– hazzey
Sep 4 '19 at 16:01

You miss a piece of key information: the "beginning" temperature of the room. Oh, and for the love of God, what is wrong with SI units that you Americans never use them?

Converting units

10"=0.254m
17"=0.4318m
397F=202.778C=475.778K


Ok, from converting units makes me wonder the sanity of yours and your teacher. But considering that this is an Engineering SE, and we Engineers are never known to be sane in the first place...

Assumption 1: The heater is just a single point - A room barely reaches 0.5m (just around your thigh level) will not be enough to put any regular heater in. Therefore, just assume this for a calculation exercise like this

Assumption 2: Humidity is ignore

Assumption 3: Ideal gas

Assumption 4: Begin temperature as normal, or 25C, 298K

The temperature change is $$\Delta T=475.778-298=177.778K$$

The heat capacity of air: according to this site, the heat capacity changes according to temperature. The same goes for density from a reference source

But, writing them in full would be long and tidious (and I doubt such is the case for a simple question like this), so just assume the 2 values would not be changed, therefore, $$c=1.003$$ and $$\rho =1.225$$

The mass of the air is $$m=\rho * V = 1.225*0.254*0.254*0.4318=0.034126kg$$

The total heat energy required (assumption number 6: No heat loss) is $$Q=m*c*\Delta T=0.034126*1.003*177.778=6.085kJ$$

For one hour heating, the net heat output power is

$$P=Q/t=6.085/3600=1.69*10^{-3}kW=1.63W$$

So, this is significantly smaller than the heat output of 2kW of the heater. Something is wrong, either with my series of assumption, or with the way this question is given.

• Nice work and +1 for conversion to metric. Some of the British colonies still haven't caught up. Tips: SE supports HTML entities in the posts (but not in the comments). &deg;, &Omega;, &mu;, &deg;, &times;, etc. as well as <sup>...</sup> and <sub>...</sub>. For your MathJAX you can use \times rather than * and, strictly speaking, your units should be separated from the numbers by a space and non-italicised. e.g. 6.085\ \text{kJ}. For fractions use \frac {top}{bottom}. Sep 4 '19 at 17:39
• Thanks. I'm still fairly new to SE (I register an account long ago, but I only aggressively use it recently). Any comment on how to use MathJax is greatly appreciated. Sep 5 '19 at 1:07