100 g of ice at 0 ºC is added to an insulated chamber containing 20 g of steam at 100 ºC. What is the final temperature of the 120 g of water

Question

I began by saying the following:

Q_{absorbed by ice}=Q_{released by steam}

⇒ Q₁ + Q₂ = Q₃ + Q₄

∴ m_iceL_f + m_icec_wΔT = m_steamL_v + m_steamc_wΔT

m_iceL_f + m_icec_w(T_f - T_i) = m_steamL_v + m_steamc_w(T_f - T_i)

Where did I go wrong? Do I make the right side of the equation negative?

Note:

Answer 1

You did everything right, only for ice (T_t - T_i) = T_f - 273K, for steam you have to take (T_i - T_f) = (373K - T_f)