Answer:
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
Explanation:
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
We have mass of copper = m = 25.3 g
Specific heat of copper = c = 0.385 J/g°C
ΔT = 39°C - 22°C = 17°C
Heat absorbed by the copper :
The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.
Answer:
This question is incomplete
Explanation:
This question is incomplete because the result of the described experiment would have better determined the type of scientific explanation to profer. However, the type of material that will preserve the relative hotness or temperature of the hot coffee for the longest time will be a material than can resist heat transfer. These materials tend to keep hot substances hot by not allowing the heat of the coffee to be conducted or pass through it. These materials are mostly insulators or made by placing an insulator between two heat conductors.
Generally, heat is usually transferred from a region of higher concentration to a region of lower concentration, hence when the heat is denied of this transfer, the heat will remain trapped in the "heat-donor" substance (in this case the hot coffee). Thus, the material chosen (A, B or C) will be the material that resists heat transfer the most based on the explanation above.
Answer:
It's C, direct and peripheral.
Explanation:
Just took the test
Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
