How many grams of water (s=1.00 cal/g C) will release 1367 J of heat when cooled from 45.2 C to 36.2 C? a) 42 g
b) 36 g
c) 31 g
d) 28 g

Mass of water  = 36 g

Explanation:

Given data:

Amount of energy released = 1367 J = 1367/4.184 = 327 cal

Initial temperature = 45.2 °C

Final temperature = 36.2 °C

Amount of water = ?

Solution:

Specific heat capacity of water is 1 cal/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

ΔT = 36.2 °C -  45.2 °C

ΔT = -9°C

-327 cal = m×1 cal/g.°C ×  -9°C

m = -327 cal/-9cal/g

m = 36 g

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