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

the answer is 303.15 or 303