Question

17.
200 g of hot water at 80°C is added to 300 g of cold water at 10°C. Calculate the
final temperature of the mixture of water. Consider the heat taken by the container
to be negligible. (Take, specific heat capacity of water is 4200 J kg-10c-1)
​

Answer 1

Answer:

Given,

Specific heat of water SW=4.2 kJ/kg K

Mass of cold, water,MC=0.3kg

Mass of hot, water, MH=0.2kg

If final temperature is Tf

Heat Loos = heat gain

MHSW(80−Tf)=MCSW(Tf−10)

Tf=MHSW+MCSWMHSW×80+MCSW×10

Tf=0.2×4.2+0.3×4.20.2×4.2×80+0.3×4.2×10

Tf=38oC

Final temperature of mixture is 38

Answer 2

Answer:

ok

Explanation:

here is the answer

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