Question

Practice Problems 5 :
1. A solid of mass 150 g at 200°C is placed in 0.4 kg of water at 20°C till a constant temperature is attained. If
the S.H.C. of the solid is 0.5 JgKfind the resulting temperature of the mixture.
127.7 °C)​

Answer 1

Explanation:

m= 150j-015kg

1=200-/0

s¹=0.5jg-1kg_1=500jkg/=?

from water

1²=10-20 s²=4200jk-¹k-¹10=?

from low of consevation energy ( heat lost by body =heat gained by water )

so

m1 s1 /1=m2 s2 /2

0.15 ×500 ( 200- /0) = 0.4 ×4200 ( 10- 20)

75 ( 200- 10 ) = 1680 ( 10 - 20)

1500- /5 /0=1680/0-33

1605/5/0 = 18600 600

11.5 ⁰c

here heat absorbed to raise the term of ice from to (100⁰c

equation 1 = sm◇1

2100×0.04 × 10= 8405 heat abosver to rain term of water for 0⁰c t0 80 ⁰c

at = sm ◇1 42.00× 1+80 - 336000j

total heat absorbed =840+336000=336840j

Answer 2

the resulting temperature of the mixture = -1,2

i hope this will help you .ヽ(;^o^ヽ)