The outpost structure of the bacteria cell is the.

64.71C

Explanation:

Given

Mass (m) = 5g

Initial Temperature (T1) = 25C

Heat (Q) = 832 joules

Required

Determine the final temperature

This question will be solved using:

Q = mc∆T

Where c = specific heat = 4.19

Substitute values in the above formula, we have:

832 = 5 * 4.19 * (T2 - 25)

832 = 20.95 * (T2 - 25)

Divide both sides by 20.95

39.71 = T2 - 25

Add 25 to both sides

39.71 + 25 = T2

T2 = 64.71C