Answer:
The mass is [tex]m_w = 0.599 \ kg[/tex]
Explanation:
From the question we are told that
The mass of ice is [tex]m_c = 0.20 \ kg[/tex]
The initial temperature of the ice is [tex]T_i = -40.0 ^oC[/tex]
The initial temperature of the water is [tex]T_{iw} = 80^o C[/tex]
The final temperature of the system is [tex]T_f = 20^oC[/tex]
Generally according to the law of energy conservation,
The total heat loss is = total heat gained
Now the total heat gain is mathematically represented as
[tex]H = H_1 + H_2 + H_3[/tex]
Here [tex]H_1[/tex] is the energy required to move the ice from [tex]-40^oC \to 0^oC[/tex]
And it mathematically evaluated as
[tex]H_1 = m_c * c_c * \Delta T[/tex]
Here the specific heat of ice is [tex]c_c = 2100 \ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
So
[tex]H_1 = 0.20 * 2100 * (0-(-40))[/tex]
[tex]H_1 = 16800\ J[/tex]
[tex]H_2[/tex] is the energy to melt the ice
And it mathematically evaluated as
[tex]H_2 = m * H_L[/tex]
The latent heat of fusion of ice is [tex]H_L = 334 J/g = 334 *10^{3} J /kg[/tex]
So
[tex]H_2 = 0.20 * 334 *10^{3}[/tex]
[tex]H_2 = 66800 \ J[/tex]
[tex]H_3[/tex] is the energy to raise the melted ice to [tex]20^oC[/tex]
And it mathematically evaluated as
[tex]H_3 = m_c * c_w * \Delta T[/tex]
Here the specific heat of water is [tex]c_w= 4190\ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
[tex]H_3 = 0.20 * 4190* (20-0))[/tex]
[tex]H_3 = 16744 \ J[/tex]
So
[tex]H = 16800 + 66800 + 16744[/tex]
[tex]H = 100344\ J[/tex]
The heat loss is mathematically evaluated as
[tex]H_d = m * c_h ( 80 - 20 )[/tex]
[tex]H_d = m_w * 4190 * ( 80 - 20 )[/tex]
[tex]H_d = 167600 m_w[/tex]
So
[tex]167600 m_w = 100344[/tex]
=> [tex]m_w = 0.599 \ kg[/tex]
