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 :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

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:

$Q=15.5g\times 4.184J/g^oC\times (50.0-20.0)^oC$

$Q=1945.56J=1.94\times 10^3J=1.94kJ$

Therefore, the heat absorbed by the water is 19.4 kJ