Question

B) A 120.0 g sample of ice at -10.0°C has placed on a heat source. How much energy must be

Answer 1

Answer:

Q = 27060 j

Explanation:

Given data:

Mass of sample = 120 g

Initial temperature = -10.0 °C

Final temperature = 100 °C

Heat added = ?

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 = T₂ - T₁

ΔT = 100- (-10)

ΔT = 110°C

Q = m.c. ΔT

Q = 120 g × 2.05 g/J.°C × 110°C

Q = 27060 j