An electric range burner weighing 683.0 grams is turned off after reaching a temperature of 477.6°c, and is allowed to cool down to 23.2°c.

calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 552.0 grams of water from 23.2°c to 80.3°c.

0.102 cal/g.°C

Explanation:

According to the law of conservation of energy, the sum of the heat released by the electric burner (Qb) and the heat absorbed by the water (Qw) is zero.

Qb + Qw = 0

Qb = -Qw

Both heats can be calculated using the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

Then,

Qb = -Qw

cb × mb × ΔTb = - cw × mw × ΔTw

cb = - cw × mw × ΔTw / mb × ΔTb

cb = - (1 cal/g.°C) × 552.0 g × (80.3°C - 23.2°C) / 683.0 g × (23.2°C - 477.6°C)

cb = 0.102 cal/g.°C

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.