he initial and final temperatures of a liquid placed on a heater are r

1.	he initial and final temperatures of a liquid placed on a heater are recorded as (30.6±0.2) °C and (68.3±0.1) °C respectively. Find the rise in temperature. (37.6±0.3) °C (37.7±0.1) °C (37.7±0.3) °C (37.7±0.2) °C
Answer» HLO MATE here is ur ans..... It is given that, INITIAL TEMPERATURE= (40.6±0.2)° C Final temperature= (78.3±0.3)° C Now, rise in temperature=Final temperature-Initial temperature. So, =(78.3±0.3)-(40.6±0.2) On rearranging the terms, =(78.3-40.6)±(0.2+0.3) =(37.7)±(0.5) =(37.7±0.5)° C Answer=> (37.7±0.5)° C..... hope this hepuh dear..... pls mark as brainliest

he initial and final temperatures of a liquid placed on a heater are recorded as (30.6±0.2) °C and (68.3±0.1) °C respectively. Find the rise in temperature. (37.6±0.3) °C (37.7±0.1) °C (37.7±0.3) °C (37.7±0.2) °C

Answer»

HLO MATE here is ur ans.....

It is given that,

INITIAL TEMPERATURE= (40.6±0.2)° C

Final temperature= (78.3±0.3)° C

Now, rise in temperature=Final temperature-Initial temperature.

So,

=(78.3±0.3)-(40.6±0.2)

On rearranging the terms,

=(78.3-40.6)±(0.2+0.3)

=(37.7)±(0.5)

=(37.7±0.5)° C

Answer=> (37.7±0.5)° C.....

hope this hepuh dear.....

pls mark as brainliest

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he initial and final temperatures of a liquid placed on a heater are recorded as (30.6±0.2) °C and (68.3±0.1) °C respectively. Find the rise in temperature. (37.6±0.3) °C (37.7±0.1) °C (37.7±0.3) °C (37.7±0.2) °C

It is given that,

INITIAL TEMPERATURE= (40.6±0.2)° C

Final temperature= (78.3±0.3)° C

Now, rise in temperature=Final temperature-Initial temperature.

So,

=(78.3±0.3)-(40.6±0.2)

On rearranging the terms,

=(78.3-40.6)±(0.2+0.3)

=(37.7)±(0.5)

=(37.7±0.5)° C

Answer=> (37.7±0.5)° C.....

hope this hepuh dear.....

pls mark as brainliest

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