Manipal Entrance Test 2013 Physics Paper

Form Einstein's mass energy relation
E=Δmc2 .....(i)
where Δm is mass lost, c is speed of light. Also heat given by body is
E=mCΔθ .....(ii)
where, C is specific heat
Equating Eqs (i) and (ii) we get
Δm=

MCΔθ

C2

=

M×0.2×100×4.2×103

(3×108)2

J

Δm

M

=

20×4.2×103

(3×108)2

% increase in mass
=

Δm

M

×100=

20×4.2×103

(3×108)2

×100
=9.3×10−11%