If 420 joules of heat energy is added to 25 grams of water at 25 degrees celcius, what will be the final temperature of the water, in Celcius degrees?

T2 = 29°C

Explanation:

Given data:

Heat added = 420 j

Mass of water = 25 g

Initial temperature = 25°C

Final temperature = ?

Solution;

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water = 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values.

420 j = 25 g ×4.18 j/g.°C × (Final temperature - initial temperature)

420 j = 25 g ×4.18 j/g.°C × (T2 - 25°C)

420 j = 104.5  j/°C × (T2 - 25°C)

420 j /104.5  j/°C = T2 - 25°C

4°C + 25°C = T2

T2 = 29°C

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

hurricanes form when warm, moist air from the ocean surface quickly rises into the atmosphere and forms storm clouds and rain. when the clouds are formed, heat is released which warms the cool air in the upper regions of the atmosphere so that the cycle can continue. which two phases of the water cycle are repeated in this process?