Question

How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degrees F? (The vegetables have a 60% water content and the specific heat of solid vegetables is 0.24 BTU/lb F) Heat = Number BTU

Answer 1

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

`Q =mc\Delta T`

where,

m = mass

c = specific heat

`\Delta T` = Change in temperature

Therefore the total heat exchange is given as

`\Delta Q = Q_w+Q_v`

`\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)`

Our values are given as,

Total mass is
`M_T` = 200lb ,however the mass of solid vegetable and water is given as,

`m_v= 0.4*200lb = 80lb`

`m_w=0.6*200lb=120lb`

`T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF`

Replacing at our equation we have,

`\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)`

`\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)`

`\Delta Q = 22411.2Btu`

Therefore the heat removed is 22411.2 Btu