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ivolga24 [154]
3 years ago
12

A Н20= C=1 h=4 o=4 2. H3PO4 + КОН KзРО І

Physics
1 answer:
maks197457 [2]3 years ago
7 0

Answer:

Н20= C=1 h=4 o=4

Explanation:

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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, i
abruzzese [7]

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

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