A cooking vessel on a slow burner contains 10.0 kg of water and an unknown mass of ice in equili

A cooking vessel on a slow burner contains 10.0 kg of water...

A cooking vessel on a slow burner contains 10.0 $\mathrm{kg}$ of water and an unknown mass of ice in equilibrium at $0^{\circ} \mathrm{C}$ at time $t=0 .$ The temperature of the mixture is measured at various times, and the result is plotted in Figure $\mathrm{P} 20.65 .$ During the first 50.0 $\mathrm{min}$ , the temperature increases to $2.00^{\circ} \mathrm{C}$ . Ignoring the heat capacity of the vessel, determine the initial mass of ice.

Key Concepts

Phase Equilibrium

Phase equilibrium occurs when different phases of a substance, such as ice and water at 0°C, coexist in a stable balance. Understanding this equilibrium is important when analyzing systems where a substance is undergoing a phase change, since the temperature remains constant until the entire phase transition is complete.

Latent Heat of Fusion

This concept refers to the amount of energy required to change a substance from the solid phase to the liquid phase at its melting point, without changing its temperature. In problems involving ice melting in water, it determines how much heat energy is absorbed by the ice during the phase transition.

Specific Heat Capacity

This is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is essential in calculating the energy transfer needed to change the temperature of substances, such as water, after any phase change has occurred.

Energy Conservation

The principle of energy conservation in thermodynamics ensures that the total energy in a closed system remains constant. It is crucial in setting up the energy balance equations, equating the heat lost by one component to the heat gained by another, to determine unknown quantities like mass or temperature changes.

Transcript

00:01 So here in this problem we have to find the initial mass of the ice for the given condition and we are given that a cooking vessel contains mass of water of a 10 kg and an unknown mass of ice and let the unknown mass of ice be m i that we have to calculate and both of them are at temperature 0 degrees celsius at time t equals to zero now the temperature of the temperature of the temperature of the temperature of the mixture rises as the time increases and we are given a plot also for the temperature the time graph so let us draw that graph first so this is the graph given for temperature time graph and at this time this is 50 minutes from 50 to 60 minutes the temperature of the mixture rises by 2 degrees celsius you can observe in the graph also so let us start with the solution so we know that in this period from 0 to 50 the ice is melting because temperature is not increasing that means ice will be melting and we can apply the conservation of energy that power supplied times the multiplied with time it will be equals to it will be equals to heat lost by the ice so p for this period time is 50 minutes so we can write 50 minutes here and q is m ice em of ice into latent heat of fusion and we know at the latent heat of fusion in case of ice it is 3 .33 into 10 to the power 5 joule per g and mi we have to find out so we can write this equation as p into 50 minutes is equals to m i times 3 .33 into 10 to the power 5 joule per g so now i'll be considering this case when the time is from 50 minute to 60 minute and in this case also the energy goes into the system at a constant rate p so again by using the conservation of energy we can write that p into delta t is equal to q and q in this case will be mc delta t so mass here is both the we have to take the mass of ice also and the water also that means m of water water plus m of ice into a specific heat and specific heat is you know it is 4186 it's 6 jule per kg degree celsius and the change in temperature is 2 degrees celsius so it will be equals to p delta t here is 10 minutes so 10 minutes and mass of water we have already given it's 10 kg so we can write 10 kg here so from here after simplification we'll get pay into 10 minutes is equals to 83 .7 kilojoule plus 8 .37 kilojoule per kg times m i and m i is a mass of ice that we have to find out so from here we can write at or we can do one thing okay let me do this way only that from here p is equals to m i times 3 .33 into 10 to the power 5 joule per g divided by 50 minutes so we can plug the value of p in this equation this one so when i plug p in this we get m .m.

04:15 Times 3 .33 into 10 to the power 5 jule per kg and it's divided by 50 minutes and is multiplied by this 10 minute this as it is and it will be equals to 83 .7 kilojoule plus 8 .37 kilojoule per kg times m i so here this 10 minute and 5 15 minutes will cancelled out and we are left in the denominator as 5 from here if we simplify this we'll get mi is equals to 83 .7 kilojoule divide by 66 .6 minus 8 .37 it's kilojoule per kg and after dividing the two terms we'll get 1 .44 kg so finally we can conclude that the unknown mass of ice was 1 .44 kg...

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