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 =2100 ~J ~kg ⁻¹ ~K ⁻¹ Specific heat capacity of water =4186 ~J ~kg ⁻¹ ~K ⁻¹ Latent heat of fusion of ice =3.35 1

Question

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 =2100 ~J ~kg ⁻¹ ~K ⁻¹ Specific heat capacity of water =4186 ~J ~kg ⁻¹ ~K ⁻¹ Latent heat of fusion of ice =3.35 10^5 ~J ~kg ⁻¹

Quantrex Academy · Accepted Answer

Correct answer: A. 96316 ~J

We have Q = mS _ i [-10^ -0^ ]- mL _ f + mS _ w [0^ -30^ ] aligned & =- mS _ i (+10^ )- mL _ f - mS _ w (30^ ) & =-0.2 2100 10-0.2 3.35 10^5-0.2 4186 30 & =-96316 ~J aligned -ve sign indicate that heat is absorbed So, Q_ absorbed =96316 ~J

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 =2100 ~J ~kg ⁻¹ ~K ⁻¹ Specific heat capacity of water =4186 ~J ~kg ⁻¹ ~K ⁻¹ Latent heat of fusion of ice =3.35 10^5 ~J ~kg ⁻¹

Options

Correct answer

Step-by-step solution

More from Thermal Properties of Matter