Answer:
yes? is there a question...
Explanation:
Answer:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ
If by Period you mean Column, they share the same amount of electrons in their outer shell.
Answer:
Molar enthalpy change, the enthalpy change for one mole of a pure substance
Explanation: i think
Answer:
<h3>The man's average body temp. will fall by 0.6°C to 38.4°C</h3>
Explanation:
The enthalpy (heat content) of the water, using a datum of 0°C, is
Hw = Mw kg x Cp,w (specific heat capacity kJ/kg °C) x Tw °C
= 1 kg x 4.18 kJ/kg°C x 3°C = 12.5 kJ
Hman (pre drink) = 68 kg x 3.6 kJ/kg/°C x 39°C = 9547 kJ
So Hman (post drink) = H (pre drink) + Hw = 9547 + 12.5 = 9559.5 kJ because no heat is lost immediately.
But Hman (after drink) has mass 68 + 1 = 69 kg
Also his new Cp will be approx (3.6 x 68/69) + (4.18 x 1/69) = 3.608 kJ/kg°C
So Hman = 9559.5 kJ (from above) = 69 kg x 3.608 kJ/kg°C x Tnew
Therefore the man's new overall temp. = 38.4°C which is a drop of 0.6°C
