Question

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, leaving only the water. Find the final temperature $T_f$ of the water. The freezing point of water is 0°C.

Answer 1

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, $L=335000\ J.kg^{-1}$

Now according to the given that the whole ice melts:

using energy conservation with heat,

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

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

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

$T_f=68.808\ ^{\circ}C$

Answer 2

The final temperature of water is $T_f =68.808^oC$

Given:

mass of water,  $m_w=0.50kg$

initial temperature of water, $T_{wi}=80^oC$

mass of ice, $m_i=0.050kg$

initial temperature of ice, $T_{ii}=-5^oC$

we know, Specific heat capacity of water, $c_w=4186J.kg^{-1}.K^{-1}$

Specific heat capacity of ice, $c_i=2100 J.kg^{-1}K^{-1}$

Latent heat of fusion of ice, $L=335000J.kg^{-1}$

It is given in the question that after a while, all the ice melts, leaving only the water.

Heat is a form of energy, heat is conserved, i.e., it cannot be created or destroyed. It can, however, be transferred from one place to another.

Using energy conservation with heat,

$m_i.c_i(T_f-T_i)+m_iL+m_w.c_w(T_f-T_i)=0\\\\m_i.c_i(T_f-T_i)+m_iL= - m_w.c_w(T_f-T_i)\\\\0.05*2100*T_f-(-5)+ 0.05*335000= -0.5*4186*(80-T_f)\\\\T_f= 68.808^oC$

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