Ahunk of up gold weighs 800 gram to heat the gold up to a temperature of 200 kelvin if the gold absorbs 12 kilojoules of energy what was the initial temperature of the gold? the specific heat capacity of gold is 0.129 grams times c

The initial temperature of the gold was 84 K.

Step 1. Calculate the temperature change.

The formula for the heat (q) absorbed is

q = mCΔT

We can solve this equation to get

ΔT = q/(mC)

q = 12 000 J; m = 800 g; C = 0.129 J·K⁻¹g⁻¹

∴ ΔT = 12 000 J/(800 g × 0.129 J·K⁻¹g⁻¹) = 116 K

Step 2. Calculate the initial temperature

ΔT = T_f – T_i = 200 K – T_i = 116 K

T_i = 200 K – 116 K = 84 K

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.