Question

What 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

Q=MCdeltaT
*make sure you change celcius into Kelvin*
Q=20(4.18)(20)
Q=1,672

It requires 1,672 Joules

Answer 2

Answer : The amount of heat required is, 1672 J

Solution :

Formula used :

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

where,

Q = heat required = ?

m = mass of water = 20 g

c = specific heat of water =
`4.18J/g^oC`

`\Delta T=\text{Change in temperature}`

`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 amount of heat required.

`Q=20g* 4.18J/g^oC* (30-10)^oC`

`Q=1672J`

Therefore, the amount of heat required is, 1672 J