Question

3. How much ice must be added to 100 g water at 30ºC in order to reduce its temperature to 20ºC? (A) 10 g (B) 80 g (C) 400 g (D) None of these

Answer 1

Answer:

[Heat capacity of water is Cpw = 4.186J/g-K]

[Heat of fusion of ice Hi = 333.55J/g]

Mass of ice is m grams;

To melt the ice at 0C (assumed) to water at 0C requires

ΔH1 = mHi = m(333.55) = 333.55m J

To raise the temperature of this water from 0C to 20C requires

ΔH2 = mCpwΔT = m(4.186)(20 – 0) = 83.72m J

To cool 100g of the water from 30C to 20C releases

ΔH3 = mCpwΔT = 100(4.186)(30 – 20) = 4186J

But ΔH1 + ΔH2 = ΔH3, so

333.55m + 83.72m = 4186

m = 4186/417.27 = 10.03

So, the mass of ice required is 10g

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Answer 2

$hey \: mate \: here \: is \: your \: answer.$

Answer:

The answer is 10 g

Explanation:

Let the mass of ice be x

So, x×80+x×20=1000[100×10×1=1000car]

⇒x=10g

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