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lapo4ka [179]
3 years ago
8

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

ure of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC.
Physics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, T_i=4^{\circ} C

Final temperature of water, T_f=25^{\circ} C

The specific heat capacity of liquid water is, c=4.184\ J/g\ ^oC

Heat transferred is given by :

Q=mc(T_f-T_i)

Q=710\times 4.184\times (25-4)

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

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The acceleration object D undergoes is 25 m/s².

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