Question

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 water 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

Answer 1

the answer is 5.27*10^4 (52,700)

Answer 2

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