Question

How many joules are needed to warm 45.0 grams of water from 30.0 degrees C to 75.0 degrees C?

Answer 1

q=m x Cp x ∆T

m is the mass of the water, Cp is the specific heat of water and ∆T is the change in temperature of the water (final-initial temperature). q is the energy involved in the reaction, measured in joules.

q=(45.0) x (4.184 Jg^-1/°C^-1) x (45°C)
q=8472.6 Joules

Answer 2

Answer : The amount of heat needed are, 8464.5 J

Explanation :

Formula used :

`q=m* c* (T_(final)-T_(initial))`

where,

q = heat gained = ?

m = mass of water = 45.0 g

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

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

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

Now put all the given values in the above formula, we get:

`q=45.0g* 4.18J/g^oC* (75.0-30.0)^oC`

`q=8464.5J`

Thus, the amount of heat needed are, 8464.5 J