To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.
By definition the exchange of heat is given by
[tex]Q =mc\Delta T[/tex]
where,
m = mass
c = specific heat
[tex]\Delta T[/tex] = Change in temperature
Therefore the total heat exchange is given as
[tex]\Delta Q = Q_w+Q_v[/tex]
[tex]\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)[/tex]
Our values are given as,
Total mass is [tex]M_T[/tex] = 200lb ,however the mass of solid vegetable and water is given as,
[tex]m_v= 0.4*200lb = 80lb[/tex]
[tex]m_w=0.6*200lb=120lb[/tex]
[tex]T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF[/tex]
Replacing at our equation we have,
[tex]\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)[/tex]
[tex]\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)[/tex]
[tex]\Delta Q = 22411.2Btu[/tex]
Therefore the heat removed is 22411.2 Btu
