Question

Thermal decomposition of a rail car load of limestone to lime and carbon dioxide requires 2.97 x 106 kJ of heat.

Convert this energy to calories.

Once you have the value, determine its log (base 10).

Answer 1

Answer: a)  $68.3\times 10^6cal$

b) $log_{10}(68.3\times 10^6cal)=7.83$

Explanation:

Heat is defined as a spontaneous flow of energy from one object to another. It is measured in Joules, calories, kilo Joules etc.

These units of energy are inter convertible.

We are given:

a) Energy absorbed by limestone = $2.97\times 10^6kJ$

Converting this unit of temperature into $calories$ by using conversion factor:

1 kJ = 239.006 calories

$2.97\times 10^6kJ=\frac{239.006}{1}\times 2.97\times 10^6=68.3\times 10^6cal$

Thus the energy in calories is $68.3\times 10^6cal$

b) The value of log base 10 of $68.3\times 10^6cal$  is:

$log_{10}(68.3\times 10^6cal)=7.83$