Common brick or decorative tile used even with noncombustible backing, in general, is not sufficient thermal protection for a solid fuel-burning appliance. Please refer to the sample calculations that demonstrate this.

Thermal conductivity "k" can be specified/measured in two different ways:

In the first way "k" is measured/specified per inch of material.

In the second way, "k" is specified for the actual thickness of the material being used, even if this thickness is less than or greater than one inch.

QUESTION 1: The Manufacturer specifies that a nominal "k" value of 0.7 (BTU)/(Ft 2) (Hr) ( oF) be installed for floor protection. I can get material that has a "k" of 0.5 (BTU) (Inch)/Ft 2) (Hr) ( oF) and is available in ½ inch and 1 inch thicknesses. Which one should I install?

ANSWER 1 In this case :

  • "k" actual thickness specified by the manufacturer is known
  • "k" per inch of a material desired to be used is known and the desired material thicknesses available for purchase are known.
  • It is necessary to calculate the "actual thickness "k" value

PROBLEM SOLUTION - SOLVING FOR AN ACCEPTABLE "k" VALUE for the actual thickness intended installed

Try the use of the ½ inch product first:

0.5 (BTU) (Inch)/Ft 2) (Hr) ( oF) ÷ 0.5 inch thickness = 1.0 (BTU)/(Ft 2) (Hr) ( oF) ( actual thickness performance achieved) but since 1.0 > 0.7 required, the ½ inch thick material is unacceptable ("k" actual installed must be equal to or less than "k" actual thickness required)

Next, try the 1 inch product:

0.5 (BTU) (Inch)/Ft 2) (Hr) ( oF) ÷ 1.0 inch thickness = 0.5 (BTU)/(Ft 2) (Hr) ( oF) and

since 0.5 < 0.7 required, the 1 inch thick product of "k" = 0.5 (BTU) (Inch)/Ft 2) (Hr) ( oF) is acceptable.

QUESTION 2 The Manufacturer has specified a material with a "k" of 0.5 (per inch) be installed and that the required thickness is at least 1.0 inch. I have on hand material that has a "k" of 1.95 (per inch). If I use this material, instead of the material that the Manufacturer specifies, then what thickness should I use?

ANSWER 2 In this case:

  • The "k" value (per inch) and the thickness of a specified acceptable material is known.
  • An alternative material is desired to be used and the "k" value (per inch) of the alternative material is known.
  • It is desired to know what the required thickness of the alternative material must be in order to be equivalent to the specified acceptable material of known "k" and known thickness.


It is only necessary to divide the alternative "k" value by the specified "k" value and then multiply by the specified thickness to determine the necessary thickness of the desired alternative material as demonstrated below:

SOLUTION [ 1.95 (BTU) (Inch)/Ft 2) (Hr) ( oF) ÷ 0.5 (BTU) (Inch)/Ft 2) (Hr) ( oF)] x 1.0 inches = 3.9 inches which is the required thickness for the desired alternative material - this thickness is significantly larger than for the specified "k" and specified thickness because the "k" value of the desired alternative material is so much higher than for the specified material .

QUESTION 3: The appliance Manufacturer specifies a hearth extension material with a "k" (per inch) of 0.35 and a required thickness of 1 inch. I want to install a decorative ceramic tile bonded to an insulating board. The insulating board has a "k" (per inch) = 0.6 and is 1 inch thick. The ceramic tile is ¼ inch thick and has a "k" (per inch) of 13.5

ANSWER 13 In this case:

  • The appliance Manufacture specifies a specific value of "k" (per inch) and provides a specified thickness of the floor protection system.
  • It is desired to know what thickness and combination of materials (forming a composite floor protection system) will be necessary.


The appliance Manufacturer specifies a hearth extension "k" (per inch) of 0.35 and a required thickness of 1 inch. Instead we want to use a composite of a ceramic tile, bonded to an insulating board. Given the thermal properties as stated above, how thick must the insulating backer board be? Important: note the high thermal conductivity of the ceramic tile. It is a relatively poor insulating material.

1. First convert "k" to "R" {to convert "k" to "R"; "R" = [actual inches of thickness] ÷ "k" (per inch)]}

"R" specified = 1.0 inch ÷ 0.35 (BTU) (Inch)/Ft 2) (Hr) ( oF) = 2.86 ( oF) (Hr) (Ft 2) / Btu

"R" backer board = 1.0 inch ÷ 0.6 (BTU) (Inch)/Ft 2) (Hr) ( oF) = 1.67 ( oF) (Hr) (Ft 2) / Btu

"R" ceramic tile = 0.25 inch ÷ 13.5 (BTU) (Inch)/ Ft 2) (Hr) ( oF) = 0.019 ( oF) (Hr) (Ft 2) / Btu

Noting that with composite materials one simply adds "R" values, the total "R" value of ceramic tile bonded to a single backer board sheet is:

1.67 + 0.019 = 1.69 ( oF) (Hr) (Ft 2) / Btu < the specified "R" value of 2.86 ( oF) (Hr) (Ft 2) / Btu thus a single backer board + ceramic tile system has too low an "R" value.

SOLUTION - Add a second backer board of "R" = 1.67 ( oF) (Hr) (Ft 2) / Btu

so that the total "R" of the desired composite of insulating backer board and decorative ceramic tile is: "R" total = "R" backer board w/tile + "R" backer board

"R" total = 1.69 + 1.67 = 3.36 ( oF) (Hr) (Ft 2) / Btu > 2.86 therefore ok.

Important Note: A combustible material cannot be sandwiched between the ceramic tile and the insulating board. In this case the ceramic tile needs to be bonded directly to the insulating board. If the ceramic tile were bonded to plywood, that is sandwiched between the tile and the insulating board, then the "k' of the composite is only as good as the tile, which will have insufficient thermal protection.