Answer:
The mass of ice to be added is [tex]M =0.085 \ kg[/tex]
Explanation:
From the question we are told that
The mass of water is [tex]m_w = 0.300 \ kg[/tex]
The temperature of water is [tex]T_w = 71.5 6^oC = 344.6 \ K[/tex]
The specific heat of water is [tex]c_w = 4190 J / kg \cdot K[/tex]
The specific heat of ice is [tex]c_i = 2100 J/kg \cdot K[/tex]
The heat of fusion of water is [tex]H_f =334 kJ/kg= 3.34*10^{5} J/kg[/tex]
The final temperature is [tex]T_f = 40^oC = 313 \ K[/tex]
The temperature of ice is [tex]T_i = 252.1 \ K[/tex]
Generally according to the law of energy conservation
The heat gained by ice = The heat lost by water
The heat lost by water is mathematically represented as
[tex]Q_w = m_w * c_w * (T_w - T_f)[/tex]
substituting values
[tex]Q_w = 0.300 * 4190 * (344.5 - 313)[/tex]
[tex]Q_w = 39595.5 J[/tex]
The amount of heat gained by the ice is
[tex]Q = Q_i + Q_f + Q_l[/tex]
Where [tex]Q_i[/tex] is the amount of heat absorbed by the ice to get to [tex]0^o C[/tex] (273 K)and this is mathematically represented as
[tex]Q_i = M * c_i * (273 - T_i)[/tex]
Here M is the mass of ice
substituting values
[tex]Q_i = M * 2100 * (273 - 252.1)[/tex]
[tex]Q_i = 43890M[/tex]
[tex]Q_f[/tex] is the heat gained by the ice as it is been converted to liquid
[tex]Q_f = M * H_f[/tex]
substituting values
[tex]Q_f = 3.34 *10^{5 } M[/tex]
And [tex]Q_l[/tex] is the heat gained by the ice(now water) as it is been heated to [tex]40^oC[/tex] and it is mathematically represented as
[tex]Q_l = M * c_w * (T_f -273)[/tex]
substituting values
[tex]Q_l = M * 2190 * (313 -273)[/tex]
[tex]Q_l = 87600 M \ J[/tex]
So
[tex]Q = 43890 M + 3.34 *10^5 M + 87600M[/tex]
[tex]Q = 465490 M \ J[/tex]
From the law of energy conservation
[tex]39595.5 = 465490 M[/tex]
=> [tex]M =0.085 \ kg[/tex]
