Ice at -20,^oC is added to 50,g of water at 40,^oC. When temperature of mixture reaches 0,^oC, it is

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Ice at $-20\,^oC$ is added to $50\,g$ of water at $40\,^oC.$ When the temperature of the mixture reaches $0\,^oC,$ it is found that $20\,g$ of ice is still unmelted. The amount of ice added to the water was close to ........$g$ (Specific heat of water $= 4.2\,J/g/^oC)$ Heat of fusion of water at $0^oC = 334\,J/g$ )

A

$50$

B

$100$

C

$60$

D

$40$

(JEE MAIN-2019)

Solution

$m \times 0.5 \times 20 + \left( {m – 20} \right) \times 80$

$ = 50 \times 1 \times 40$

$ \Rightarrow 90\,m – 1600 = 2000$

$ \Rightarrow 90\,m = 3600$

$ \Rightarrow m = 40\,gm$

Standard 11

Physics

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