Question

Specefic heat of lead is 120 . When 7200 J of heat is supplied to 5kg of lead , The rise in temprature is .

Answer 1

Heya.......!!!!

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This question can be easily solved by applying the formulae
:- Q = ms∆t

Here -

• Q = 7200

• m = 5 Kg

• s = 120

Putting the values , we get ,,

∆t = Q/ms

=>
$\frac{7200}{5 \times 120} \\ \\ = > \: 12$

∆t. => 12℃


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Answer 2

Q=MS. deltaT
DELTA T=Q/M*S=7200/120*5
HENCE DELTA T=12
Here is your answer
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