Two metals A and B lose the same quantity of heat when their temperatures drop from 20°C to 15°C. If the specific heat Capacity of A Ii thrice that of B, calculate the ratio of the mass of A to that of B.
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The correct answer is E
Change in temperature=20-15=5
Therefore;
A=B
From heat change,
H=mc×change in temperature
H=H
Therefore
M×C×5=3×M×C×5
Dividing both sides by C
3C=15C
1C=3C
Therefore;
1:3
The selected answer is wrong:
Heat lost by both is same,therefore, MaCa¥=MbCb¥
(¥,stands for change in temperature)
Ca=heat capacity of A=3Cb
therefore, Ma3Cb=MbCb
3Ma=1Mb
d ratio of Ma to Mb is1:3.
Ie Ma/Mb=1/3
the correct answer is a because a is three times that of b and since b is 5 then a is 5x3 the it is 15 : 5 thats 3 :1
To find the ratio of the masses of metals A and B, we use the fact that they lose the same quantity of heat when their temperatures drop from 20°C to 15°C. The specific heat capacity of A is given as three times that of B.
Let’s denote:
𝑐
𝐴
c
A
as the specific heat capacity of metal A
𝑐
𝐵
c
B
as the specific heat capacity of metal B
𝑚
𝐴
m
A
as the mass of metal A
𝑚
𝐵
m
B
as the mass of metal B
Δ
𝑇
ΔT as the change in temperature, which is the same for both metals
We know:
𝑐
𝐴
=
3
𝑐
𝐵
c
A
=3c
B
The heat lost by A and B is the same, so:
𝑄
=
𝑚
𝐴
⋅
𝑐
𝐴
⋅
Δ
𝑇
=
𝑚
𝐵
⋅
𝑐
𝐵
⋅
Δ
𝑇
Q=m
A
⋅c
A
⋅ΔT=m
B
⋅c
B
⋅ΔT
Since
Δ
𝑇
ΔT is the same for both metals, it cancels out:
𝑚
𝐴
⋅
𝑐
𝐴
=
𝑚
𝐵
⋅
𝑐
𝐵
m
A
⋅c
A
=m
B
⋅c
B
Substituting
𝑐
𝐴
=
3
𝑐
𝐵
c
A
=3c
B
:
𝑚
𝐴
⋅
3
𝑐
𝐵
=
𝑚
𝐵
⋅
𝑐
𝐵
m
A
⋅3c
B
=m
B
⋅c
B
Dividing both sides by
𝑐
𝐵
c
B
:
𝑚
𝐴
⋅
3
=
𝑚
𝐵
m
A
⋅3=m
B
Solving for the ratio of the masses:
𝑚
𝐴
𝑚
𝐵
=
1
3
m
B
m
A
=
3
1
So the ratio of the mass of A to that of B is
1
:
3
1:3.
Therefore, the correct answer is:
E. 1:3
