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One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is 125 kgand its initial temperature is20°C, (a) how much energy must the water transfer to its surroundings in order to freeze completely and (b) what is the lowest possible temperature of the water and its surroundings until that happens?

Short Answer

Expert verified
  1. The energy that the water must transfer to its surroundings in order to freeze completely is5.2×107J
  2. The lowest possible temperature of the water and its surroundings until that happens is0°C

Step by step solution

01

The given data

  1. Mass of the water,m=125 â¶Ä‰k²µ
  2. Initial temperature,Ti=200C
  3. Specific heat capacity of water,c=4190 J/°ì²µ.°­
  4. Specific latent heat of water,LF=333×103J/kg
02

Understanding the concept of specific heat

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. We can use the concept of specific heat of the water and change of state. The heat transformation involved in the phase change from liquid to solid is called heat of fusion.

Formulae:

The heat energy required by a body,Q=³¾³¦Î”°Õ …(¾±)

Where, m = mass

c = specific heat capacity

Δ°Õ= change in temperature

Q= required heat energy

The heat energy released or absorbed by the body,Q=Lm …(¾±¾±)

Where, m = mass

L = specific latent heat

03

(a) Calculation of the energy transfer by the water

The energy transferred by the water to the surrounding in order to freeze completely:

The initial temperature of the water is 20°C. In order to freeze the water, energy is released by water in two steps. First, the released energy decreases the temperature from20°Cto 0°C, therefore, the energy transferred by the water is given as:

Q1=cm(Tf−Ti)

When the water freezes completely, then there is phase change and the energy transferred by the water isQ2=LFm

Hence, the total energy transferred to the surroundings is given as:

Q=Q1+Q2=4190J/kg.K×125kg×(20°C−0°C)+333×103J/kg×125kg=5.2×107J

Hence, the value of the required energy is5.2×107J

04

(b) Calculation of the lowest temperature of water

The lowest temperature of the water before all the water freezes is given as:TLow=0°C

Hence, the value of the lowest temperature is0°C

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