How many Joules are released to cool 250.0 grams of liquid water from 100°C to 0°C? The specific heat of water is 4.180 J/g.C.
Is this endothermic or exothermic?

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Starrysoul100 Starrysoul100 · Answer 1 · 2022-02-11T14:11:34+01:00

[tex]\bold{\huge{\orange{\underline{ Solution}}}}[/tex]

[tex]\bold{\underline{ Given :- }}[/tex]

We have 250g of liquid water and it needs to be cool at temperature from 100° C to 0° C
Specific heat of water is 4.180J/g°C

[tex]\bold{\underline{ To \: Find :- }}[/tex]

[tex]\bold{\underline{ Let's \:Begin:- }}[/tex]

We know that,

Amount of heat energy = mass * specific heat * change in temperature

That is,

[tex]\sf{\red{ Q = mcΔT }}[/tex]

Subsitute the required values in the above formula :-

[tex]\sf{ Q = 250 × 4.180 ×(0 - 100 )}[/tex]

[tex]\sf{ Q = 250 × 4.180 × - 100 }[/tex]

[tex]\sf{ Q = 250 × - 418}[/tex]

[tex]\sf{\pink{ Q = - 104,500 J }}[/tex]

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

[tex]\bold{\underline{ Now :- }}[/tex]

We have to tell whether the above process is endothermic or exothermic :-

Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic