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If the specific heat of water is 4.18J/g°C and 325J of heat is added to 56.75g of water, howmuch does the temperature of the water change by?3253 46EeT5gl9oC137

Question
If the specific heat of water is 4.18J/g°C and 325J of heat is added to 56.75g of water, how
much does the temperature of the water change by?
3253 46
EeT5gl9oC
137
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If the specific heat of water is 4.18J/g°C and 325J of heat is added to 56.75g of water, how much does the temperature of the water change by? 3253 46 EeT5gl9oC 137

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Step 1

We have to calculate the temperature change when 325 J of heat is added to 56.75 g of water.

Specific heat of water = 4.18 J/g°C.

Step 2

Heat is the energy that is transferred from the hotter system to the cooler system and it can be expressed in terms of specific heat capacity (C), mass of the sample and the initial and final temperature of the system.

q =mxC xAT
where
qamount of heat energy gained or lost
m mass of the sample
C specific heat of the sample
AT change in temperature
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q =mxC xAT where qamount of heat energy gained or lost m mass of the sample C specific heat of the sample AT change in temperature

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Step 3

Given-

mwater =56.75 g

Cwater = 4.18 J/g°C

q = 325 J

Therefore, temperature chang...

x Δ .
q=mx C
325 J-(56.75 g) x (4. 18 J/g'C) xΔΤ
325 J
=1.37 C
.. ΔΤ-
(56.75 g) x (4.18 J/g C)
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x Δ . q=mx C 325 J-(56.75 g) x (4. 18 J/g'C) xΔΤ 325 J =1.37 C .. ΔΤ- (56.75 g) x (4.18 J/g C)

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