if the temperature of 800.0 g h2O increases by 45 degree c how much heat (in joules and in calories) was added to the water?

Amount of heat added 36.02 cal.

Amount of heat added 150696 j.

Explanation:

Given dа ta:

Mass of water = 800 g

Change in temperature = 45 °C

Amount of heat added = ?

Solution:

Formula:

Q = m. c. ΔT

Q = amount of heat released or added

m = mass of substance

c = specific heat of water

ΔT = change in temperature

Now we will put the values in formula.

Q = m. c. ΔT

Q = 800 g * 4.186 j / g. °C * 45 °C

Q = 150,696 j

In calories,

150696 / 4184 = 36.02 cal