A 10.0-kg block of gold is heated in the Sun from 25.0℃ to 30.0℃. Use the table (Specific heat table) to help calculate the change in the block's thermal energy.

900 J
1300 J
6400 J
3900 J

Q = 6450 j

Explanation:

Given data:

Mass of gold = 10 kg (10×1000 =10,000 g)

Initial temperature = 25°C

Final temperature = 30°C

Energy absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of gold = 0.129 j/g.°C

ΔT = 30°C - 25°C

ΔT = 5°C

Q = 10,000 g ×0.129 j/g.°C× 5°C

Q = 6450 j

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.