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To find the specific heat capacity of a solid of mass 600 g whose temperature was 40oC, it was placed in a calorimeter that contains 25 g of water 10oC. The mixture reached a final temperature of 25oC. How much is the metal's specific heat capacity? (Cwater = 4180 J/Kg.K)


500J/Kg.K

250J/Kg.K

174J/Kg.K

2000J/Kg.k






Answer and I will give you brainiliest
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Sagot :

Spectrojamzidn

Answer:

m of solid = 600g = 0.6kg

T=40°C

It was placed in a calorimeter that contains 25g(0.025kg) of water at 10°C

Now. We would have considered the calorimeter in this solving because it will be at the same temp as the water which it is holding. Example... If you put a hot water inside a Cup... The cup itself will become hot and have the same temp as the water you put in it ...

But in this question... We'll Ignore the Calorimeter and the heat gained by it since no parameter was given for it and also We don't know the material which the calorimeter is made of.

So

Since the Solid is at a higher temp(40°)... It will lose heat to water; which is at a lower temp.

From the question... THE EQUILIBRIUM TEMPERATURE REACHED IS 25°C

Then

Temp Change for solid(heat loss) = 40-25=15°C

Temp Change for water(heat gain)= 25-10 =15°C

Heat lost by solid = Heat gained by water

MC∆T = MC∆T

0.6 x C x 15 = 0.025 x 4180 x 15

C= 0.025x4180x15÷(0.6x15)

C= 174.16 ~ 174J/Kg.K.

OPTION C IS LEGIT

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