The specific heat of water = 4200J kg ^ – 1 K ^ – 1 and the latent heat of ice = 3.4 × 10 ^ 5 J k g ^ – 1 . 100 grams of ice at 0 ^ o C is placed in 200g of water at 25 ^ o C . The amount of ice that will melt as the

Correct answer: A. 61.7 MS Δ T = Ml (200)/(1000) × 4200 × 25 = m × 340 × 10 ^ 3 m = 61 . 7

JEE Main 20204 SeptemberMorning Shift

JEE Main 2020 · 4 September · Morning Shift · Physics · Thermal Properties of Matter

The specific heat of water = 4200J kg ^ – 1 K ^ – 1 and the latent...

The specific heat of water

= 4200\text{J}

{\text{kg}}^{– 1}{\text{K}}^{– 1}

and the latent heat of ice

= 3.4 \times {10}^{5}\text{J}

{\text{g}}^{– 1}

100

grams of ice at

0{}^{\text{o}}\text{C}

is placed in

200\text{g}

of water at

25{}^{\text{o}}\text{C}

. The amount of ice that will melt as the temperature of water reaches

0{}^{\text{o}}\text{C}

is close to (in grams)

A. $61.7$

$MS \Delta T = Ml$

$\dfrac{200}{1000} \times 4200 \times 25 = m \times 340 \times {10}^{3}$

$m = 61 . 7$