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11. Calculate the amount of heat transferred when 550 grams of water cools from an initial temperature of
25.0 °C to a final temperature of 4.0 °C. The specific heat capacity of liquid water is 4.184J/g °C. Please
express you answer as a number in units of k), and be careful about the sign (positive or negative number?)
since that shows the direction of heat flow.

11 Calculate the amount of heat transferred when 550 grams of water cools from an initial temperature of 250 C to a final temperature of 40 C The specific heat class=

Respuesta :

Eduard22sly

Answer:

–48.3 KJ

Explanation:

Data obtained from the question. This includes the following:

Mass (M) = 550g

Initial temperature (T1) = 25°C

Final temperature (T2) = 4°C

Change in temperature (ΔT) = T2 – T1 = 4°C – 25°C = –21°C

Specific heat capacity (C) = 4.184J/g°C

Heat (Q) =.?

The heat transferred can be obtained as follow:

Q = MCΔT

Q = 550 x 4.184 x –21

Q = – 48325.2 J = –48.3KJ

Therefore, the heat transferred is –48.3KJ since we are cooling the water.

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