Answer:
The substance has a specific heat of 1.176 J/g°C
Explanation:
<u>Step 1: </u>Data given
Temperature change = 34 °C
Mass of the substance = 20 kg = 20000 grams
The substance gained 800 kJ of heat during this temperature change
<u>Step 2:</u> Calculate the specific heat
q = m*c*ΔT
⇒ with q = heat gained = 800 kJ = 800000 J
⇒ with m = the mass of the substance = 20 kg = 20000 grams
⇒ with c = the specific heat of the substance = TO BE DETERMINED
⇒ with ΔT = the change of temperature = T2 -T1 = 48° - 14 ° = 34°
c = q/(m*ΔT)
c = 800000 / (20000 * 34)
c = 1.176 J/g°C
The substance has a specific heat of 1.176 J/g°C
Answer:
your answer would be : examples such as Thallium-206, Lead-210, & Bismuth-214.
<span>the formation of a gas
</span>
Density of a liquid determines how it will layer (heaviest to lightest). If the liquid is least dense it will float to the bottom. Layers will remain separated because each liquid is actually floating on top of the more dense liquid beneath it.
P = pressure, V = volume, T = absolute temperature k = constant.