Question

10g of water at 10 degree celsius is mixed with 50 g of copper at 10 degree celsius ,find the final temperature of mixture when specific heat capacity of water is 4.2J/g degree celsius and of copper is 0.4J/g celsius
Please give correct solution ,please

Answer 1

Answer:

If 10 g of ice at - 10° C is added to 50 g of water at 15° C, what is the temperature of the mixture?

SBI Magnum Children’s Benefit Fund - investment plan.

We have ice at -10°C,

For the ice to come to 0°C, the energy required is m*s*∆T

m is mass of ice = 10g

s is specific heat of ice = 0.8 cal/g.°C

∆T is temperature difference = 0-(-10) = 10°C

So the energy required is 10*0.8*10 = 80 cal

The energy released when 50 grams of water at 15°C becomes water at 0°C is m*s*∆T

s for water is 1 cal/g.°C

Energy released is 50*1*(15–0) = 50*15 = 750 cal

For the ice to completely turn into water at 0°C, the energy required is m*L

L is latent heat of ice = 80 cal/g

Energy required is 10*80 = 800 cal.

But the water when turns into water at 0°C, the energy released is only 750 cal.

So, the total ice will not get converted into water.

The energy left after the ice gets to 0°C is 750–80=680 cal

With 680 cal, the amount of ice that turns into water is 680/80 = 8.5 grams.

So the composition is 10–8.5=1.5 grams of ice at 0°C and 10+8.5=18.5 grams of water at 0°C.

Explanation:

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