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Chapter 3: Q53E (page 577)

An insulated beaker with negligible mass contains of 0.250kg water at 75.0°C.How many kilograms of ice -20.0°Cat must be dropped into the water to make the final temperature of the system 40.0°C ?

Short Answer

Expert verified

The mass of ice is 0.0675kg

Given: Mass of water is mwater=0.250kgtemperature of water Twater=75°C, temperature of ice is Tice=-20°Cand final

temperature of system is Tfinal=40°C

Step by step solution

01

The conservation of heat energy

The heat gain by gain of ice is equals to the heat lost by water

Qice=-Qwater

02

Heat gain by ice and heat lost by water

The heat gain by ice can be calculated as

Qice=MiceCice∆T+MiceLice+MiceCwater∆T

The lost by water can be calculated as

Qwater=MwaterCwater∆T

03

Calculation of mass of iceUsing

Qice=-QwaterMiceCice∆T+MiceLice+MiceCwater∆T=MwaterCwater∆TMice=MwaterCwater∆TwaterCice∆T+Lice+Cwater∆T

Where, Latent heat of fusion of ice is Lice=33.6×103J/KandCwater,Ciceare the specific heat capacity of water and ice whose values are4.19J/kg.k,2.108J/kg.krespectively.

Now, putting the values of constants in above equation

Mice=0.2504.19×103×4.19×10340°C-75°C2.108×1030°C--20°C+33.6×103+4.19×10340°C-0°CMice=0.0675kg

Thus, the mass of ice is 0.0675kg

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