01270 747 008 (UK)

# Heater Power & Temperature Calculations

The calculator below will help you to determine the effects of heating power on different materials. It can be useful for finding how hot something will get when heated with a heater of a certain power level, or for working out how much heater power you need to reach a certain temperature.

The calculations should only be used as a guide as in the real world there are many more variables not accounted for here. In the list of materials, the heat capacity values are given for when it is at 25 celsius. With most materials the heat capacity of the material will change as it heats up. If you want to account for this, then select “other" as the material and then input the heat capacity manually.

We have also included an input for power loss to help account for heat lost to the surroundings by a hot object. Since the shape and size of an object will significantly affect how much heat it radiates to the surroundings, you will need to estimate a value for power loss.

There are three calculations available below, simply click one of the questions to display the relevant form.

### How hot will something get?

 Material (Cp) Choose the material being heated Heat Capacity (Cp) Enter the material heat capacity in J/g°C J/g°C Mass Amount of matter in grams g Heater Power Heating power in watts W Time No. of seconds heat is applied s Loss Power lost as heat to surroundings W How hot will {mass}g of {material} get in {time} seconds when heated with a {heat}W heater? Formulae: Q = m x Cp x ΔT ΔT = Q ÷ m ÷ Cp Working: ({heat}W - {pd}W) x {time}s = {output1}J {output1}J ÷ {mass}g ÷ {cp}J/g°C = {output} Result: {mass}g of {material} will increase in temperature by {output} when heated with a {heat}W heater for {time} second(s).

### How long will it take for something to reach a specific temperature?

 Material (Cp) Choose the material being heated Heat Capacity (Cp) Enter the material heat capacity in J/g°C J/g°C Mass Amount of matter in grams g Heater Power Heating power in watts W Temperature Change Target temperature in °C °C Loss Power lost as heat to surroundings W How long will it take to heat {mass}g of {material} by {temp}°C with a {heat}W heater? Formula: Q = m x Cp x ΔT t = Q ÷ (P - PD) Working: {mass}g x {cp}J/g°C x {temp}°C = {output1}J {output1}J ÷ ({heat}W - {pd}W) = {output} Result: It will take {output} to heat {mass}g of {material} by {temp}°C with a {heat}W heater.

### How much heater power is needed to heat something to a certain temperature?

 Material (Cp) Choose the material being heated Heat Capacity (Cp) Enter the material heat capacity in J/g°C J/g°C Mass Amount of matter in grams g Time No. of seconds heat is applied s Temperature Change Target temperature in °C °C Loss Power lost as heat to surroundings W Heater power required to heat {mass}g of {material} change {temp}°C in {time} secs Formula: Q = m x Cp x ΔT P = (Q ÷ t) + PD Working: {mass}g x {cp}J/g°C x {temp}°C = {output1}J ({output1}J ÷ {time}s) + {pd}W = {output} Result: A heater power of {output} is required to heat the material by {temp}°C in {time} seconds.

### Real World Examples

1. Heating a steel part in an induction heater

This example uses our CRO-1 Induction Heater Circuit, and our CT-400 Induction Heater Coil to heat a 2.2g M6 steel nut. Using a 15V supply, the system pulls about 4.5A of current which is equivalent to 67.5W of heating power. Measurement of the temperature shows after 30s it is at around 78°C (Temperature increase of 58°C). To get a simlar result from the calculator a power loss of 65.5W is assumed. The majority of this power loss would come from the hot metal part dissipating heat to the surrounding air. Measuring again at 180 seconds, the temperature is now 237°C which would mean the power loss to the air is now up to around 66.25W.