Answer:
[tex]a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g[/tex]
Explanation:
First we need to state our assumptions:
Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice [tex]h_i_f=333.7KJkg[/tex]
Mass of water,[tex]m_w=\rho V =1\times0.3=0.3Kg[/tex].
Energy balance for the ice-water system is defined as
[tex]E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w[/tex]
a.The mass of ice at [tex]0\textdegree C[/tex] is defined as:
[tex][mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g[/tex]
b.Mass of ice at [tex]20\textdegree C[/tex] is defined as:
[tex][mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g[/tex]
c.Mass of cooled water at [tex]T_c_w=0\textdegree C[/tex]
[tex]\bigtriangleup U_c_w+\bigtriangleup U_w=0[/tex]
[tex][mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g[/tex]
