The amount of heat needed to heat 200 grams of ice at -10^{\circ} {C} to convert it into water at 30^{\circ} {C} is Specific heat capacity of ice =2100 {Jkg}^{-1} {K}^{-1} Specific heat capacity of water =4186 {J} {kg}^{-1} {K}^{-1} Latent heat of fusion of ice =3.35 \times 10^{5} {J} {kg}^{-1}

AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Physics

Question : 97 of 160

Marks: +1, -0

The amount of heat needed to heat 200 grams of ice at −10∘C to convert it into water at 30∘C is
Specific heat capacity of ice =2100Jkg−1K−1
Specific heat capacity of water =4186Jkg−1K−1
Latent heat of fusion of ice =3.35×105Jkg−1

96316J
67000J
92116J
71200T

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