Answer to Question #173640 in Chemistry for Monte

Question

A chemist heats a 215-g sample of iron from 25°C to 90°C. How much heat (q) did the iron absorb? The specific heat capacity of iron is 0.45 J/g°C. Use q = mcΔT

Answer 1

Solution:

q = m × C × ΔT

q = m × C × (T_f - T_i),

where:

q = amount of heat energy gained or lost by substance (J)

m = mass of sample (g)

C = specific heat capacity (J °C^-1 g^-1)

T_f = final temperature (°C)

T_i = initial temperature (°C)

 

Thus:

q = (215 g) × (0.45 J ^oC^-1 g^-1) × (90 - 25)°C = 6288.75 J

q = 6288.75 J = 6.3 kJ

Answer: the iron absorb 6.3 kJ of heat (q).