Question

Hot water of mass 7 kg at 98°C is cooled for taking bath by mixing 14 kg of water at 10°C. Calculate the final temperature of water. Specific heat capacity of water is 4200J/kg°C. Neglect the heat taken by the bucket. ​

Answer 1

Answer:

Let the final temperature of the mixture be

t

∘

C

Equating the heat loss and heat gained we get

20

×

4.2

×

(

t

−

20

)

=

10

×

4.2

×

(

90

−

t

)

⇒

20

t

−

400

=

900

−

10

t

⇒

30

t

=

1300

⇒

t

=

43.3

∘

C

Answer 2

Answer:

Specific heat capacity of water=4200 J/kgK

By the principle of mixtures,

m x c (T1-T) = m x c x(T-T2)

2 x 4200(80-T) = 8 x 4200(T-25)

(4200 on both sides get cancelled)

160-2T = 8T-200

360 = 10T

T= 36 degree Celcius