a piece of copper weighing 400g is heated to 100degree celsius and then quickly transfered to a copper calorimeter of mass 10g containing 100g of liquid of unknown specific heat capacity at 30 degree celsius.if the final temperature of the mixture is 50 degree celsius,calculate the specific heat capacity of the liquid
The formula is H=ML
L=H/M
WHERE M (ICE) =M3-M2
H= MC (@1-@2)
M= WATER ( M1-M2)
41.96g
Given;
M1= 400g
M2= 10g
M3= 100g
@1= 100
@2= 30
@3= 50
C=
H= M*L
L= H/M
M3-M2
@1-@2
M2-M1
Please send formula
Thanks♡︎
Write out the parameters
And check c(SHC) with c =C/M
To calculate the specific heat capacity of the liquid, we can use the principle of energy conservation.
The heat gained by the copper can be calculated using the formula:
Qcopper = mcopper * c * ΔT
where,
mcopper = mass of the copper = 400g
c = specific heat capacity of copper = 0.39 J/g°C (given)
ΔT = change in temperature = final temperature - initial temperature = 50°C - 100°C = -50°C
Now, since heat is transferred from the copper to the liquid, the heat gained by the liquid is equal to the heat lost by the copper:
Qcopper = Qliquid
So,
mcopper * c * ΔT = mliquid * cliquid * ΔT
where,
mliquid = mass of the liquid = 100g
ΔT = change in temperature = final temperature - initial temperature = 50°C - 30°C = 20°C
cliquid = specific heat capacity of the liquid (what we need to find)
Rearranging the equation, we have:
cliquid = (mcopper * c * ΔT) / (mliquid * ΔT)
Substituting the given values:
cliquid = (400g * 0.39 J/g°C * -50°C) / (100g * 20°C)
Simplifying the equation:
cliquid = -390 J/g°C / 20°C
cliquid = -19.5 J/g°C
We find that the specific heat capacity of the liquid is approximately -19.5 J/g°C.
Note: The negative sign indicates that the liquid is losing heat in this process.
sum of heats added is zero.
Heat added to liquid+heataddedtocopper=0
100*C*(50-30)+400*Ccu*(50-100)=0
look up specific heat Ccu of copper, and solve for C
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