In the past, restaurants had four hours, straight through, to cool food to 41°F or lower. Now the FDA recommends cooling food in two stages -- from 135°F to 70°F in two hours then from 70°F to 41°F or lower in an additional four hours for a total cooling time of six hours
Explanation:
the two-stage cooling method<span> is a </span><span>Food Code </span>counselled<span> procedure for cooling food in restaurants and foodservice </span>institutions<span>. </span>within the<span> two-stage cooling </span>methodology<span>, food is</span><span> cooled from 140° F (60° C) to 70° F (21° C) </span>among 2<span> hours and to 41° F (5° C) or lower </span>among<span> four hours. Use of this cooling </span>methodology<span> ensures that food is cooled quickly and safely and has no harmful effects.</span>
Answer:
• The closing of the frontier
Explanation:
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Answer:
The correct option is: <u>B. 366 torr</u>
Explanation:
Given: <u>On the ground</u>- Initial Volume: V₁ = 8.00 m³, Initial Atmospheric Pressure: P₁= 768 torr;
<u>At 4200 m height</u>- Final Volume: V₂ = 16.80 m³, Final Atmospheric Pressure: P₂ = ?
Amount of gas: n, and Temperature: T = constant
<u>According to the Boyle's Law</u>, for a given amount of gas at constant temperature: P₁ V₁ = P₂ V₂
⇒ P₂ = P₁ V₁ ÷ V₂
⇒ P₂ = [(768 torr) × (8.00 m³)] ÷ (16.80 m³)
⇒ P₂ = 365.71 torr ≈ 366 torr
<u>Therefore, the final air pressure at 4200 m height: P₂ = 366 torr.</u>
Answer:
c = 0.13 j/ g.°C
Explanation:
Given data:
Mass of mercury = 29.5 g
Initial temperature = 32°C
Final temperature = 161°C
Heat absorbed = 499.2 j
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 - T1
ΔT = 161°C - 32°C
ΔT = 129 °C
Q = m.c. ΔT
c = Q / m. ΔT
c = 499.2 j / 29.5 g. 129 °C
c = 499.2 j / 3805.5 g. °C
c = 0.13 j/ g.°C
Answer:
measures earth or ground movement such as earthquake