Answer:
c = 0.07 j/g.k
Explanation:
Given data:
Mass of sample = 35 g
Heat absorbed = 48 j
Initial temperature = 293 K
Final temperature = 313 K
Specific heat of substance = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = Final temperature - initial temperature
ΔT = 313 k - 293 K
ΔT = 20 k
Now we will put the values in formula.
48 j = 35 g × c× 20 k
48 j = 700 g.k ×c
c = 48 j/700 g.k
c = 0.07 j/g.k
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Answer:
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Explanation:
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To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
