Question

A Certain freezing process requires the room
temperature to be lowered from 60oC at
he roteet of 8'covery hour what will
be the room tempor
e room temperature 8 hours after
the process begins?​

Answer 1

$\huge \mathfrak \red{answer}$

$\sf{initial \: temperature = 60°c}$

$\sf{room \: temperature \: deceases \: 8°c \: per \: hour}$

$\sf \red{temperature \: change \: in \: one \: hour}$

$\sf \red{ = - 8°c}$

$\sf \red{temperature \: change \: in \: 8hrs}$

$\sf \red{ - 8 \times 8°c}$

$\sf \red{ = - 64°c}$

room temperature after 8 hours

$\sf{initial \: temperature + temp \: changes \: in \: 8hrs}$

$\rm \red{60+ ( - 64)}$

$\sf \red{ = 60 - 64}$

$\sf \red{ = ( - 60 + 64)}$

$\sf \red{ = 64 - 60}$

$\sf \red{4c}$

room temperature after 8 hours 4°c

Answer 2

Answer:

\sf{initial \: temperature = 60°c}initialtemperature=60°c

\sf{room \: temperature \: deceases \: 8°c \: per \: hour}roomtemperaturedeceases8°cperhour

\sf \red{temperature \: change \: in \: one \: hour}temperaturechangeinonehour

\sf \red{ = - 8°c}=−8°c

\sf \red{temperature \: change \: in \: 8hrs}temperaturechangein8hrs

\sf \red{ - 8 \times 8°c}−8×8°c

\sf \red{ = - 64°c}=−64°c

room temperature after 8 hours

\sf{initial \: temperature + temp \: changes \: in \: 8hrs}initialtemperature+tempchangesin8hrs

\rm \red{60+ ( - 64)}60+(−64)

\sf \red{ = 60 - 64}=60−64

\sf \red{ = ( - 60 + 64)}=(−60+64)

\sf \red{ = 64 - 60}=64−60

\sf \red{4c}4c

room temperature after 8 hours 4°c

Step-by-step explanation:

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