
When m grams of water at $$10^\circ C$$ is mixed with m grams of water at $$0^\circ C$$, which of the following statements are false?
A) The temperature of the system will be given by the equation, $m \times 8$ + $m \times 1 \times (T - 0)$ = $m \times 1 \times (10 - T)$.
B) Whole of ice will melt and the temperature will be more than $$0^\circ C$$ but lesser than $$10^\circ C$$.
C) Whole of ice will melt and the temperature will be $$0^\circ C$$.
D) Whole of ice will not melt and the temperature will be $$0^\circ C$$.
Answer
232.8k+ views
Hint: When water is added to ice then ice absorbs the heat from the water. This hidden heat is called heat of fusion and heat capacity. We have to calculate the amount of heat of fusion of ice and the heat capacity of water. If the heat of fusion is more than ice will not melt whereas if the heat capacity of water is more, then ice will melt.
Complete step by step solution:
According to thermodynamic studies, some of the calculated values we need:
Heat of fusion of ice = $333.55J/g$ and
Heat capacity of water = $4.186JK/g$
First of all, we have to find the amount of heat energy released by cooling M g of water at $$10^\circ C$$ to $$0^\circ C$$ to melt M g of ice. We get that the amount of heat can be calculated with the help of the formula $$\Delta H = mC\Delta T$$. So for 10g of water the amount of heat will be 41.86M Joules of heat. Now the amount of heat of ice fusion is 333.55j/g. Therefore, the ice will not melt and the temperature remains the same i.e. $$0^\circ C$$.
For melting of ice, it should be given heat more than its heat of fusion. Heat of fusion is defined as the amount of heat required to extract from water to convert into ice. And the heat capacity is defined as the amount of heat in unit mass of water. So to convert a unit mass of water into ice, this heat has to be removed from the water to bring it at $$0^\circ C$$. And to convert it into ice heat equal to heat of fusion has to be removed from it.
Note: To convert substances from one state to another latent heat of fusion has to be given or extracted from the substances. Also every substance has heat capacity at specific temperature. So firstly the substance has to be brought to $$0^\circ C$$. Then latent heat of fusion has to be removed in order to convert water into ice.
Complete step by step solution:
According to thermodynamic studies, some of the calculated values we need:
Heat of fusion of ice = $333.55J/g$ and
Heat capacity of water = $4.186JK/g$
First of all, we have to find the amount of heat energy released by cooling M g of water at $$10^\circ C$$ to $$0^\circ C$$ to melt M g of ice. We get that the amount of heat can be calculated with the help of the formula $$\Delta H = mC\Delta T$$. So for 10g of water the amount of heat will be 41.86M Joules of heat. Now the amount of heat of ice fusion is 333.55j/g. Therefore, the ice will not melt and the temperature remains the same i.e. $$0^\circ C$$.
For melting of ice, it should be given heat more than its heat of fusion. Heat of fusion is defined as the amount of heat required to extract from water to convert into ice. And the heat capacity is defined as the amount of heat in unit mass of water. So to convert a unit mass of water into ice, this heat has to be removed from the water to bring it at $$0^\circ C$$. And to convert it into ice heat equal to heat of fusion has to be removed from it.
Note: To convert substances from one state to another latent heat of fusion has to be given or extracted from the substances. Also every substance has heat capacity at specific temperature. So firstly the substance has to be brought to $$0^\circ C$$. Then latent heat of fusion has to be removed in order to convert water into ice.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

