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dolphi86 [110]
3 years ago
7

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

Chemistry
1 answer:
MatroZZZ [7]3 years ago
5 0

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

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