Answer:
16.32 °C
Explanation:
We are given;
Mass of aluminum bowl; m_b = 0.25 kg
Mass of soup; m_s = 0.8 kg
Thus, formula to find the amount of heat energy for a temperature change of 27.6°C to 0°C is;
Q = (m_b•c_b•Δt) + (m_s•c_s•Δt)
Where;
c_b = 0.215 kcal/(kg•°C)
c_s = 1 kcal/(kg•°C)
ΔT = 27.6 - 0 = 27.6°C
Thus;
Q = (0.25 × 0.215 × 27.6) + (0.8 × 1 × 27.6)
Q = 23.5635 Kcal
Now, the energy that exits to be used to freeze the soup is;
Q' = 424 kJ - Q
Let's convert 424 KJ to Kcal
424 KJ = 424/4.184 Kcal = 101.3384 Kcal
Thus;
Q' = 101.3384 - 23.5635
Q' = 77.7749 Kcal
Amount of heat that's removed is given by;
Q_f = Q' - mL
Where;
m = m_s = 0.8 kg
L = 79.8 kcal/kg
Thus;
Q_f = 77.7749 - (0.8 × 79.8)
Q_f = 13.9349 Kcal
Then final temperature will be;
T_f = Q_f/((m_b•c_b) + (m_s•c_s))
T_f = 13.9349/((0.25 × 0.215) + (0.8 × 1))
T_f = 16.32 °C
Answer:
an infective agent that typically consists of a nucleic acid molecule in a protein coat, is too small to be seen by light microscopy, and is able to multiply only within the living cells of a host.
"a virus infection"
Explanation:
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