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A 45gpiece of ice at 0Cis added to a sample of water at 8.0C. All of the ice melts and the temperature of the water decreases to 0.0C. How many grams of water were in the sample?

Short Answer

Expert verified

450gof water in the sample when ice melts added and the temperature of water decreases to0.0C

Step by step solution

01

Given Information.

The following information's are used to find the gram of water in sample:

Mass of water=45g

Initial Temperature=8C

freezing temperature=0C

02

Energy Balance.

By using energy balance between ice and water,

Ice absorbs the heat energy lost by water. Water loses heat energy, which causes it to cool, whereas ice absorbs heat energy, which causes it to melt.

Therefore,

+miceλ=-mwaterCpΔTwater

where,

miceis mass ice, λis latent heat, mwateris mass of water, Cpis specific heat of water and ΔTwateris change in the temperature of water.

A positive sign indicates that heat energy is being gained, whereas a negative sign indicates that heat energy is being lost.

03

Find the gram of water.

Water temperature vary from 8Cto 0C

ΔTwater=Tfinal-Tinitial

=0-8C=-8C

We known,

Latent heat, λ=334J/g

Specific heat of water, Cp=4.18J/gC

Substitute the values,+mice(λ)=-mwaterCpΔTwater

localid="1651902308199" +45(g)×334(J/g)=-mwater×4.18J/g°C×-8C

localid="1651902311257" 15030J=mwater×33.44J/gmwater=15030J33.44J/g

localid="1651902314703" =449.462g=450g

Therefore, the mass of water islocalid="1651902318964" 450g

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