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

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

Solution :

Formula used :

`Q= m* c* \Delta T`

`Q= m* c* (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)* (4.18 J/g^oC)* (30-10)^oC`

Q = 1672 Joules

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

Answer 2

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