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elena-14-01-66 [18.8K]
3 years ago
7

In an insulated container, 0.50 kg of water at 80°C is mixed with 0.050 kg of ice at −5.0°C. After a while, all the ice melts, l

eaving only the water. Find the final temperature T_f of the water. The freezing point of water is 0°C.
Physics
2 answers:
Phoenix [80]3 years ago
6 0

Explanation:

When ice tends to absorb heat energy then it will change into liquid state. Now,  change in the temperature of cold water dT = T_{f} - 0^{o}C = T_{f}

Change in the temperature of original water in the container dT' = 80 -T_{f
}

Latent heat of melting of ice L = 80 cal/g

Specific heat of water C = 1 cal/g C

Heat absorbed by ice to melt = mass(m_{i}) x latent heat(L)

Heat absorbed by water = mass(m_{w}) x specific heat(C) x change in temperature (dT')

m_{i} = 0.050 kg = 50 g and m_{w} = 0.50 kg = 500 g

Now, heat lost by water is as follows.

heat lost by hot water( at 80^{o}C)  = heat absorbed by the ice + heat absorbed by cold water

         m_{i} \times L + m_{i} \times C dT = m_{w} \times C dT'

            50 \times 80 + 50 \times 1 \times (T_{f}) = 500 \times 1 \times (80 - T_{f})

So, we will cancel out 50 from each term given above. Hence,

              80 + T_{f} = 800 - 10T_{f}  

               11 T_{f} = 720

               T_{f} = 65.45^{o}C

Thus, we can conclude that final temperature of the water is 65.45^{o}C.

Morgarella [4.7K]3 years ago
5 0

Answer:

T_f=68.808\ ^{\circ}C

Explanation:

Given:

  • mass of water, m_w=0.5\ kg
  • initial temperature of water, T_{wi}=80^{\circ}C
  • mass of ice, m_i=0.05\ kg
  • initial temperature of ice, T_{ii}=-5^{\circ}C
  • Specific heat capacity of water, c_w=4186\ J.kg^{-1}.K^{-1}
  • Specific heat capacity of ice, c_i=2100\ J.kg^{-1}.K^{-1}
  • Latent heat of fusion of ice, L=335000\ J.kg^{-1}

<u>Now according to the given that the whole ice melts:</u>

using energy conservation with heat,

m_i.c_i.(T_f-T_i)+m_i.L+m_w.c_w.(T_f-T_i)=0

m_i.c_i.(T_f-T_i)+m_i.L=m_w.c_w.(T_i-T_f)

0.05\times 2100\times (T_f-(-5))+0.05\times 335000=0.5\times 4186\times (80-T_f)

T_f=68.808\ ^{\circ}C

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