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.

–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.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.