How much amount of heat is required to raise the temperature of 20 grams of water from 10°C to 30°C? The specific heat of water is 4.18 J/g°C.

Answers

Answer 1
Answer: The amount of heat:
Q = m c ΔT
m = 20 g,  c = 4.18 J/g°C,  ΔT = 30° - 10° = 20° C
Q = 20 g · 4.18 J/g°C · 20°C = 1672 J
Answer: C ) 1700 J
Answer 2
Answer:

Answer : The amount of heat is required is, 1672 J

Solution :

Formula used :

Q= m* c* \Delta T

Q= m* c* (T_(final)-T_(initial))

Q= heat gained

m= mass of the substance = 20 g

c = heat capacity of water = 4.18 J/g ° C

T_(final) = final temperature = 30^oC

T_(initial) = initial temperature = 10^oC

Now put all the given values in the above formula, we get the heat required.

Q= (20g)* (4.18 J/g^oC)* (30-10)^oC

Q = 1672 Joules

Therefore, the amount of heat is required is, 1672 J


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Answers

Answer:

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Answers

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Answers

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Answer:

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