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If 0.46 kg of water is heated from [tex]8.5^\circ C[/tex] to [tex]74.6^\circ C[/tex], the amount of heat (in kJ) absorbed by the water is (specific heat capacity of water is [tex]4.18 \, \text{J/g} \cdot \text{°C}[/tex]):

A. 102
B. 302
C. 172
D. 162
E. 127

Answer :

Final answer:

To calculate the amount of heat absorbed by the water, use the formula Q = mcΔT. The amount of heat absorbed by the water is approximately 127.03 kJ (option 'E').

Explanation:

To calculate the amount of heat absorbed by the water,

we can use the formula Q = mcΔT,

where Q is the heat absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, convert the mass of water from kg to grams: 0.46 kg = 460 g.

Then, calculate the change in temperature:

ΔT = final temperature - initial temperature = 74.6°C - 8.5°C = 66.1°C.

Finally, calculate the heat absorbed:

Q = (460 g) × (4.18 J/g°C) × (66.1°C) = 127,026.8 J = 127.03 kJ (rounded to two decimal places).

The amount of heat absorbed by the water is approximately 127.03 kJ. Hence, option 'E' is correct answer.

Learn more about Heat Absorption here:

https://brainly.com/question/12943685

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