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lianna [129]
4 years ago
9

Specific Heat of Water = 4.186

Chemistry
2 answers:
marissa [1.9K]4 years ago
5 0

Answer:

the answer is  7897.0 J

Explanation:

i did the usatestprep

yarga [219]4 years ago
4 0

Answer:

Q = 2260.44 j

Explanation:

Given data:

Mass of ice = 18 g

Initial temperature = -10 °C

Final temperature = 20°C

Heat absorbed by ice  = ?

Solution:

Specific heat capacity:

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

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 = Final temperature - initial temperature

ΔT = 20°C - (-10°C)

ΔT =  30°C

Specific heat of water = 2.00 j/g.°C

Q = 18 g × 4.186 j/g.°C × 30°C

Q = 2260.44 j

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Answer:

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