637,980 Joules is needed to raise the temperature of 2.50 kg of this oil from 23 ∘C to 191 ∘C.
<h3>What is specific heat?</h3>
The heat in calories required to raise the temperature of one gram of a substance one degree Celsius.
Q = cmΔT
where
Q = heat energy needed for that material to get desired temperature change (in Joules)
M = mass (in grams) so you have to convert from kilograms.
c = specific heat constant for the material being heated
ΔT = change in temperature
Q = (1.75)(2.17 x 1000)(191 - 23)
Q = (1.75)(2170)(168)
Q = 637,980 Joules
Hence, 637,980 Joules is needed to raise the temperature of 2.50 kg of this oil from 23 ∘C to 191 ∘C.
Learn more about the specific heat capacity here:
brainly.com/question/9560546
#SPJ1
The percentage of Calcium carbonate in chalk = 100%
<h3>Further explanation</h3>
Given
1.51 g piece of chalk produces 0.665 g of carbon dioxide
Required
percentage of calcium carbonate
Solution
Reaction
CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)
mol CO2 :
= 0.665 g : 44 g/mol
= 0.015
From the equation, mol CaCO3 = mol CO2 = 0.015
mass CaCO3 :
= mol x MW
= 0.015 x 100
= 1.5 g
Animals break down food molecules to obtain energy.
The remains of producers are broken down by decomposers.
The remains of consumers are broken down by soil decomposers.