Question

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.

Answer 1

$\bold{\huge{\orange{\underline{ Solution}}}}$

$\bold{\underline{ Given :- }}$

• 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

$\bold{\underline{ To \: Find :- }}$

• We have to find the total number of joules released.

$\bold{\underline{ Let's \:Begin:- }}$

We know that,

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

That is,

$\sf{\red{ Q = mcΔT }}$

Subsitute the required values in the above formula :-

$\sf{ Q = 250 × 4.180 ×(0 - 100 )}$

$\sf{ Q = 250 × 4.180 × - 100 }$

$\sf{ Q = 250 × - 418}$

$\sf{\pink{ Q = - 104,500 J }}$

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

$\bold{\underline{ Now :- }}$

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

• Exothermic process :- It is the process in which heat is evolved .
• Endothermic process :- It is the process in which heat is absorbed .