Question

How much amount of heat is required to raise the temperature of 20 grams of water from 10°C to 30°C? The specific heat of water is 4.18 J/g°C.

Answer 1

The amount of heat:
Q = m c ΔT
m = 20 g,  c = 4.18 J/g°C,  ΔT = 30° - 10° = 20° C
Q = 20 g · 4.18 J/g°C · 20°C = 1672 J
Answer: C ) 1700 J

Answer 2

Answer : The amount of heat is required is, 1672 J

Solution :

Formula used :

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

$Q= m\times c\times (T_{final}-T_{initial})$

Q= heat gained

m= mass of the substance = 20 g

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

$T_{final}$ = final temperature = $30^oC$

$T_{initial}$ = initial temperature = $10^oC$

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

$Q= (20g)\times (4.18 J/g^oC)\times (30-10)^oC$

Q = 1672 Joules

Therefore, the amount of heat is required is, 1672 J