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JAMB Physics 2018 Paper

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Question : 63

Total: 87

Water of mass 150g at 60

^o

c is added to 300g of water at 20

^o

c and the mixture is well stirred. Calculate the temperature of the mixture.(neglect heat losses to the surroundings)

33 $^o$ c
40 $^o$ c
25 $^o$ c
10 $^o$ c

Solution:

Water 1 => M = 150g = 0.15kg,

θ $_2$ = 60 $^o$ C; θ $_1$ = θ

Q $_1$ = MC (θ $_2$ – θ $_1$ )

= 0.15C (60 - θ)

Water 2 => M = 300g = 0.3kg

θ1 = 20 $^o$ C, θ $_2$ = θ

Q2 = MC (θ $_2$ – θ $_1$ )

= 0.3C (θ – 20)

Combining Q1 and Q2

0.15C (60 - θ) = 0.3C (θ – 20)

9 – 0.15θ = 0.3θ – 6

0.3θ + 0.15θ = 9 + 6

0.45 θ = 15

θ = 33.33 $^o$ C

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