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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

User Kidquick
by
6.2k points

1 Answer

3 votes

Answer:

Q = 6450 j

Step-by-step 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

User Michael Sync
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6.5k points