Question

How much heat is absorbed by 15.5 g of water when its temperature is increased from 20.0°C to 50.0°C? The specific heat of water is 4.184 J/(g°C).,

Answer 1

q = m . C . Δ T
q : Heat absorbed
m : mass of the sample
C : The specific heat of the substance.
Δ T : The change in temperature (Final T - initial T)
So:
q = 15.5 x 4.18 x (50 - 25) = 1619.75 J

Answer 2

Answer : The heat absorbed by the water is 19.4 kJ

Explanation :

Formula used :

`Q=m* c* \Delta T`

or,

`Q=m* c* (T_2-T_1)`

where,

Q = heat absorbed = ?

m = mass of water = 15.5 g

c = specific heat of water =
`4.184J/g^oC`

`T_1` = initial temperature =
`20.0^oC`

`T_2` = final temperature =
`50.0^oC`

Now put all the given value in the above formula, we get:

`Q=15.5g* 4.184J/g^oC* (50.0-20.0)^oC`

`Q=1945.56J=1.94* 10^3J=1.94kJ`

Therefore, the heat absorbed by the water is 19.4 kJ