Question

50 gm of hot solid of specific heat capacity .25 J per gram per °C and at 100∆°C is placed at 80 gram of cold water. When the temperature of cold water risea by 3°C. Find the initial temperature of cold water​

Answer 1

Answer:

Hot solid

M= 50g

T= 100°C

C = 0.25 =

Let 0 be the final temp.

. fall in temp. 3 (100-0)°C

Cold water at 1 °C

m=80 g

C = 4.2 =

25 100 1 4

42 10

Rise in temp. = 0 - 1 = 3 Heat lost by hot solid Heat gained by water

MCs (T-0) = mcw (0 - 1)

42 50 x (100 - 0) = 80 x x 3 10

1250 25 2 e-1008

25

e = 1250 – 1008 = 242

0 1250 x 2 25 = 19.36

But = 3

1936-1 = 3

t = 19.36 - 3 = 16.36°C

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