What is the Time Required to Boil Away 1.0kg of Ice Using 900W of Power?

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Homework Help Overview

The problem involves calculating the time required to boil away 1.0 kg of ice using a stove that produces 900 W of power. The context includes considerations of energy transfer, phase changes, and specific heat capacities.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between power, energy, and time, with some questioning the specific heat values and the concept of phase changes. There is uncertainty about the total energy required for the entire process of melting ice and boiling water.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem, including the need to break down the energy requirements into distinct phases. Some guidance has been offered regarding the concepts of specific heat and enthalpy of fusion, but no consensus has been reached on the total energy needed.

Contextual Notes

Participants note the assumption of perfect conditions (no heat loss) and express confusion about the energy calculations required for each phase transition. There is also a mention of the need to divide the problem into parts to address each energy requirement separately.

LGS1970
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Homework Statement


You have 1.0kg of ice at 0.0 deg C in a pot on a stove that produces 900 W of power. Assuming that the pot and the stove are 'perfect' (i.e., no heat lost) and ignoring the heat capacity of the pot, compute the time it will take for this ice to boil away completely.

Homework Equations


Power * time = Mass * Specific Heat * Temperature (or maybe change in temp?!)

The Attempt at a Solution


I used the equation above ^ and used the specific heat for ice of 2106 J/(kg/K) and 4218 j/(kg/K) for liquid. but I don't think this is right because what would the change in temp be for the ice to liquid phase and vapor to completely gone phase? Really confused
 
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first off do you understand that 1 watt = 1J/s ? - meaning that a stove with 1 watt produces 1 joule of energy per second.
so now forget about the stove
how much energy (joules) do you need to boil the water?

equation for change of state does not have change in temp, just becase there is no change in temperature does not mean you don't have to expend energy (joules) to make the state transition, makes sense no?

furthermore for state changes the constant is not called "specifc heat" its caled enthalpy of fusion
 
I do not know that's what the question says. I have no idea how many joules it takes to boil away completely. If i did I would have the question finished lol
 
LGS1970 said:
I do not know that's what the question says. I have no idea how many joules it takes to boil away completely. If i did I would have the question finished lol
you must divde the question into parts:
how much energy do you need to transform the water from ice to liquid;
how much energy do you need to heat said (now liquid) water from 0 to 100 degrees celsius;
and finally how much energy to transform the water from liquid to gas.

just to complete my previous answer, you use the specifc heat of ice to know the energy you need to heat ice (say from -20º to -10), you use the enthalpy of fusion to figure out how much energy you need to turn that block of ice to liquid water (once ice is at 0º), understand?
 
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