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 data:

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

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