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MaRussiya [10]
2 years ago
10

How much heat is required to increase the temperature of 198.5 grams of water from 25.0 oC to 88.5 oC? (The specific heat of wat

er is 4.18 J/g x oC)
A.52, 700 J
B.199 J
C.265 J
D.73,400 J
E.20,700 J
Chemistry
2 answers:
natali 33 [55]2 years ago
4 0

Answer : The correct option is, (A) 52700 J

Solution :

Formula used :

Q= m\times c\times \Delta T

Q = heat gained

m = mass of the substance = 198.5 g

c = heat capacity of water = 4.18 J/g°C      

\Delta T=\text{Change in temperature}=(88.5-25)^oC=63.5^oC  

Now put all the given values in the above formula, we get

Q=198.5g\times 4.18J/g^oC\times63.5^oC

Q = 52687.855 J ≈ 52700 J

Therefore, the amount of heat required is, 52700 J

arlik [135]2 years ago
3 0
The answer is 5.27*10^4 (52,700)

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