A piece of substance of specific head capacity 450JKg-1K-1 falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms-2]
2/9°C
4/9°C
9/4°C
9/2°C
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Heat energy = mc∆T
Where m =mass, c =specific heat capacity, ∆T=change in temperature
Energy at height (P.E) = mgh
From principle of energy conversion
Energy at height was converted to heat energy
Thus, mc∆T = mgh
m will cancel m
c∆T = gh
∆T = gh/c
=20x10/450
=200/450
=4/9
firstly u have to understand the relationship there; between specific heat capacity and potential energy.
therefore the equation for specific heat capacity is
Q=mcΔT
where Q= quantity of heat
m=mass in kg
ΔT= change in temperature
and for potential energy;
P. E= mgh
m=mass
g=gravity
h= height equivalent to distance
now to the calculation
DATAS; c=450
s or h=20m
g=10
ΔT=?
mcΔT=mgh
make T subject of the formula
ΔT= mgh/mc
m cancel m
so we are left with
ΔT= gh/c
ΔT= 10×20/450
ΔT=200/450
now a common sense u apply here by looking at the options
50 can go into 200, 4times
and 50 can go into 450, 9times
therefore we are left with ΔT= 4/9 final answer
energy possessed by a falling body is given as [mgh]
were m is mass,
g is acceleration due to gravity
and h to be height
heat energy is given as[mc(t1-t2)]
equating both equations we have that both masses cancel and using the resulting equation we find our answer to be 4/9 or 0.4
The distance moved by a particle in the sixth and eighth seconds of the motion are 45m and 53m respectively. Determine the acceleration and the initial speed
heat energy is given as mc(*t) : mgh i.e potentia difrenc at rest. The two masses wil cance each other leaving ct ; gh 450 * t :10* 20 then sub for t, which will give us that t : 200/450
Since d question indicate motion under gravity, hence, mc¤=mgh, where ¤=change in temperature, h=distance, thus, we substitute our values to get the correct answer.
d ques depict energy change frm P-mgh to heat MC(*t).xo Mgh=MC*t. M*10*20=M*450* *t.here,divided by coefficient of heat change(*t).u av *t=M*10*20/M*450,note dat M will cancel M. We av *t=10*20/45= 4/9
I don't just understand this solving atoll.....somebody please put me tru thanks
The question is based on conservation of energy. Therefore, c=450JKg^-1K^-1, h=20m, g=10ms^-1(delta)T "@" =change in temperature. So, mc@t=mgh, but equal masses eliminates themselves since they are in quotient(m=m). So we now have c@t=gh. Substituting, we have, 450@t=10x20 => 450@t=200, making "@t" the subject relationship, we have @t=200/450 =>4/9.
