5. In a hotel, food is stored in a freezer at -20 °C(1) What would be the temperature if it was raised by 5 °C?(ii) What would be the temperature if it was lowered by 0.5 °C every hour for 12 hours? About the author Charlotte
[tex]\boxed {\underline {\mathbb {PROPER\:QUESTION:-}}}[/tex] In a hotel, food is stored in a freezer at -20 °C, (1) What would be the temperature if it was raised by 5 °C? (2) What would be the temperature if it was lowered by 0.5 °C every hour for 12 hours? [tex]\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}[/tex] (1) [tex]\boxed{-15C}[/tex] (2) [tex]\boxed{-26C}[/tex] [tex]\boxed {\underline {\mathbb {GIVEN:-}}}[/tex] food is stored in a freezer at -20 °C (1) temperature if it was raised by 5 °C (2) the temperature if it was lowered by 0.5 °C every hour for 12 hours [tex]\boxed {\underline {\mathbb {TO\:FIND:-}}}[/tex] What would be the temperature if it was raised by 5 °C? What would be the temperature if it was lowered by 0.5 °C every hour for 12 hours? [tex]\boxed {\underline {\mathbb {SOLUTION:-}}}[/tex] Food is stored at -20°C Situation of 1 part :- It’s said that temperature is raised by 5°C [tex]\implies {+5 \: C}[/tex] [ ← increased i.e. we have taken as +5°C] Temperature when raised [tex]= -20+5\\[/tex] [tex]\boxed {\implies {-15 C}}[/tex] Situation of 2 part:- Current temperature [tex]= -20[/tex]°C Temperature lowered per hour [tex]= -0.5[/tex]°C [← lowered i.e. we have taken as -5°C] Hence temperature lowered in 12 hour [tex]= -0.5 \times 12 \\[/tex] [tex]=-6[/tex]°C Temperature after 12 hours [tex]= -20 -6[/tex] [tex]\boxed {\implies{-26C}}[/tex] ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ Reply
[tex]\boxed {\underline {\mathbb {PROPER\:QUESTION:-}}}[/tex]
In a hotel, food is stored in a freezer at -20 °C,
(1) What would be the temperature if it was raised by 5 °C?
(2) What would be the temperature if it was lowered by 0.5 °C every hour for 12 hours?
[tex]\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}[/tex]
(1) [tex]\boxed{-15C}[/tex]
(2) [tex]\boxed{-26C}[/tex]
[tex]\boxed {\underline {\mathbb {GIVEN:-}}}[/tex]
food is stored in a freezer at -20 °C
(1) temperature if it was raised by 5 °C
(2) the temperature if it was lowered by 0.5 °C every hour for 12 hours
[tex]\boxed {\underline {\mathbb {TO\:FIND:-}}}[/tex]
[tex]\boxed {\underline {\mathbb {SOLUTION:-}}}[/tex]
Food is stored at -20°C
Situation of 1 part :-
It’s said that temperature is raised by 5°C [tex]\implies {+5 \: C}[/tex] [ ← increased i.e. we have taken as +5°C]
Temperature when raised [tex]= -20+5\\[/tex]
[tex]\boxed {\implies {-15 C}}[/tex]
Situation of 2 part:-
Current temperature [tex]= -20[/tex]°C
Temperature lowered per hour [tex]= -0.5[/tex]°C [← lowered i.e. we have taken as -5°C]
Hence temperature lowered in 12 hour [tex]= -0.5 \times 12 \\[/tex] [tex]=-6[/tex]°C
Temperature after 12 hours [tex]= -20 -6[/tex]
[tex]\boxed {\implies{-26C}}[/tex]
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Answer:
Step-by-step explanation:
-20°-6° = -26°