Question

225 J of heat energy were absorbed when a 35.0 g sample of iron was warmed. What was the change in temperature?

Question 3 options:

2.89°C


3536°C


0.070°C


14.3°C

Answer 1

Answer:

14.3°C

Explanation:

q = m * c * ΔT

ΔT = $\frac{q}{mc}$

the specific heat of iron (c) is 0.450 J/g°C

ΔT = $\frac{q}{mc}$ = $\frac{225J}{35g * \frac{0.450 J}{g C} }$ = $\frac{225J * g * C}{35g * 0.450J}$

cancel out stuff now

ΔT = $\frac{225C}{35 * 0.450}$ = 14.3