Apiece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °c and dropped into 84 g of water (with an initial temperature of 20 °c) in a calorimeter. the final temperature of the system is 52.1°c. the specific heat of water is 4.184 j/g*⁰c. what is the specific heat of the metal?
a) 0.171
b) 0.343
c) 1.717
d) 3.433
The specific heat of the metal comes out to be [tex]3.423J/g^oC[/tex]
Explanation: Heat absorbed or heat lost, q is written as:
[tex]q=mc\Delta T[/tex]
We are given that an unknown metal is dropped in water. So, in this process:
Heat lost by the metal = Heat gained by the water
Mathematically,
[tex](mc\Delta T)_m=-(mc\Delta T)_w[/tex] ....(1)
For metal:
[tex]T_2=52.1^oC[/tex]
[tex]T_1=100^oC[/tex]
[tex]\Delta T=T_2-T-1=(52.1-100)^oC=-47.9^oC[/tex]
m = 68.6 g
c = ?
For Water:
[tex]T_2=52.1^oC[/tex]
[tex]T_1=20^oC[/tex]
[tex]\Delta T=T_2-T-1=(52.1-20)^oC=32.1^oC[/tex]
m = 84 g
[tex]c=4.184J/g^oC[/tex]
Putting this in equation 1, we get
[tex]68.6g\times c_m\times (-47.9^oC)=84g\times 4.184J/g^oC\times 32.1^oC[/tex]
[tex]c_m=3.423j/g^oC[/tex]
The heat absorbed by water will be equal to heat given by the metal
The heat absorbed by water = Q = msΔt
Where
m = mass of water
s = specific heat of water = 4.18 J g⁻¹°C⁻¹
Δt = final temperature - initial temperature = (52.1-20)°C = 32.1°C
So heat absorbed by water = 84 X 4.184 X 32.1 = 11281.74 J
Heat given by metal
= mass of metal X specific heat of metal X change in temperature
heat given by metal = 68.6 X specific heat of metal X (100-52.1)
11281.74 J = 68.6 X specific heat of metal X (100-52.1)
specific heat of metal = 3.43 J g⁻¹°C⁻¹
Letter b 0.343 tis the respost
52.1-4.184=47.916 You then either add or subtract this from 68.6 the first temperature of the metal. This will give you your total.
Option C. Hope I helped :)!
Option (d)
Explanation:
Let c be the specific heat of the metal.
By use of principle of caloriemetry,
Heat lost by the hot body = Heat gained by the cold body
Heat lost by metal = Heat gained by water
mass of metal x specific heat of metal x fall in temperature of metal = mass of water x specific heat of water x rise in temperature of water
68.6 x c x (100 - 52.1) = 84 x 4.184 x (52.1 - 20)
3285.94 x c = 11281.7376
c = 3.433 J/g C
The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:
-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)
where C is the specific heat capacities of the materials.
We calculate as follows:
-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 > OPTION C
Q= (m)(cp) (delta T)
cp=-q/23.8 (32.5-100.0) ( q is negative because heat was lost)