Question

How many joules of heat are absorbed to raise the temperature of 650 grams of water from 5.00c to it's boiling point, 100c

Answer 1

Answer : The amount of heat absorbed are, 258485.5 J

Solution :

Formula used :

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

where,

Q = heat gained = ?

m = mass of water = 650 g

c = specific heat of water =
`4.186J/g^oC`

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

`T_(final)` = final temperature =
`100^oC`

`T_(initial)` = initial temperature =
`5^oC`

Now put all the given values in the above formula, we get the final temperature of copper.

`Q=650g* 4.186J/g^oC* (100-5)^oC`

`Q=258485.5J`

Therefore, the amount of heat absorbed are, 258485.5 J