Answer to Question #167954 in Chemistry for Ava Brooke

Question

What is the mass of iron if 842.05 joules are added and the temperature moves from 25 ^oC to 28 ^oC?

s_Fe = 0.449 J/(gK)

Answer 1

According to the heat equation:

Q = mc(T₂-T₁)

where Q - absorbed or released heat (842.05 J), c - specific heat (0.449 J g^-1 K^-1), T₂ - final temperature (28 ^oC = 301.15 K), T₁ - initial temperature (25 ^oC = 298.15 K).

From here:

m_Fe = Q / c_Fe(T₂-T₁) = 842.05 J / [0.449 J g^-1 K^-1 × (301.15 K - 298.15 K)] = 625.13 g

Answer: 625.13 g