Question

10g of water at 80 degree is mixed with 50g of copper at 10 degree.find the final temp of the mixture .heat capacity of water 4.2j/gc and of copper is 0.4j/gc

Answer 1

The final temperature of the mixture will be 22.148°C

The mass of water = 10 g

Initial temperature of water is = 80° C

Heat capacity of water is = 4.2 J/g °C

The mass of copper = 50 g

Initial temperature of copper is = 10° C

Heat capacity of copper is = 0.4 J/g °C

Suppose, the final temperature of the mixture is = T

From the theory of heat transfer, we know,

Heat gain = Heat loss

or, Net Heat exchange between two body = constant.

Or, H = m×s×ΔT = Constant

where, m= mass, s= specific heat, ΔT= temperature change

Now,

10 × (80- T) × 4.2 = 50 × (T- 10) × 0.4

⇒ 336 - 4.2 T = 20 T - 200

⇒ 24.2 T = 536

⇒ T = 22.148

The final temperature of the mixture will be 22.148°C

Answer 2

Answer:

the final temperature will be 57.42°C

Explanation:

hot body

water

m=10g

T=80°C

c=4.2/g°C

cold body

copper

m=50g

T=10°C

c=0.4/g°C

let final temperature=T

by principal of calorimetry

heat lost by water=heat gain by copper

MCT of water=MCT of copper

10×4.2×(80-T)=50×0.4×(T-10)

10×42/10×(80-T)=50×4/10×(T-10)

42×(80-T)=20(T-10)

3360-42T=20T-200

3360+200=20T+42T

3560=62T

T=3560/62

T=57.419

T=57.42°C

So, final temperature=57.42°C