Question

0.3-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, if the ice is at (a) 0C and (b) –20C. Also (c) determine how much water would be needed if the cooling is to be done with cold water at 0C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L.

Answer 1

Answer:

$a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g$

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 $h_i_f=333.7KJkg$

Mass of water,$m_w=\rho V =1\times0.3=0.3Kg$.

Energy balance for the ice-water system is defined as

$E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w$

a.The mass of ice at $0\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g$

b.Mass of ice at $20\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g$

c.Mass of cooled water at $T_c_w=0\textdegree C$

$\bigtriangleup U_c_w+\bigtriangleup U_w=0$

$[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g$