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.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.