Question

Calculate q when 62.5 g of water is heated from 25.5°C to 80.7°C (specific heat capacity for water = 4.184J/g.K).

2.11 * 104)
6.67 * 103)
1.44 x 104)
2.78 x 104,

Answer 1

Q = 1.44×10⁴ J

Step-by-step explanation:

Given data:

Mass of water = 62.5 g

Initial temperature = 25.5°C

Final temperature = 80.7°C

Specific heat capacity of water = 4.184 J/g.K

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

ΔT = 80.7°C - 25.5°C

ΔT = 55.2 °C

Q = 62.5 g × 4.184 J/g.°C × 55.2 °C

Q = 14434.8 J

Q = 1.44×10⁴ J