Answer :
The unknown temperature of the second sample is 80 degree C
In the law of conservation of energy, energy neither be created nor be destroyed - only it change one form to another form of energy.
Now,
Mass of the first sample of water , m1 = 100 g
Initial temperature for first sample , t1 = 20 degree C
Mass of the second sample of water , m2 = 100 g
Initial temperature of the second sample , t2 = ?
Final temperature for first sample = 40 degree C
c is known as the specific heat of the water
Now we are solving this problem the law of conservation of energy
Q(gain) = - Q(lost)
(m1c(delta t)1) = -(m2(delta t)2)
m1(Final temperature- t1) = - m2(Final temperature- t2)
t2 - Final temperature = m1(Final temperature- t1) / m2
t2 - 40 = 100 (40 - 20) / 100
t2 - 40 = 20
t2 = 20 + 40
t2 = 80 degree C
Hence, the initial temperature of second sample is 80 degree C
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