Answer:
Q = 1461.6 J
Explanation:
Given data:
Mass of ice = 36 g
Initial temperature = -20°C
Final temperature = 0°C
Amount of heat absorbed = ?
Solution:
specific heat capacity of ice is 2.03 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 = T2 - T1
ΔT = 0°C - (-20°C)
ΔT = 20°C
Q = 36 g ×2.03 j/g.°C×20°C
Q = 1461.6 J
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
Answer:
it can be tooo. long or coplicated because nomatter how long u stay in school they give u homework and also a lot of people have jobs when would they be ableto work
Explanation:
545mm Hg in Kilopascals is 72.6607
I hope this helps you. Good luck stay safe, healthy and, happy!<3
You can make 10 because that is the most N2 you have. The first one that runs out limits further molecules to be made
