Question

A 0.6850 kg ice cube at -12.40°C is placed inside a rigid, thermally isolated chamber containing 6.790 moles of steam at 365.0°C. Later, you notice that the ice cube has completely melted into a puddle of water. The specific heats of ice, water, and steam are $c_{ice} = 2093$ J/(kg·°C), $c_{water} = 4186$ J/(kg·°C), and $c_{steam} = 2009$ J/(kg·°C), respectively. Calculate the final temperature $T_f$ of the puddle once it settled to equilibrium. $T_f = $ °C

          A 0.6850 kg ice cube at -12.40°C is placed inside a rigid, thermally isolated chamber containing 6.790 moles of steam at 365.0°C. Later, you notice that the ice cube has completely melted into a puddle of water.
The specific heats of ice, water, and steam are $c_{ice} = 2093$ J/(kg·°C), $c_{water} = 4186$ J/(kg·°C), and $c_{steam} = 2009$ J/(kg·°C), respectively.
Calculate the final temperature $T_f$ of the puddle once it settled to equilibrium.
$T_f = $ °C

A 0.6850 kg ice cube at -12.40°C is placed inside a rigid, thermally isolated chamber containing 6.790 moles of steam at 365.0°C. Later, you notice that the ice cube has completely melted into a puddle of water.
The specific heats of ice, water, and steam are cice = 2093 J/(kg·°C), cwater = 4186 J/(kg·°C), and csteam = 2009 J/(kg·°C), respectively.
Calculate the final temperature Tf of the puddle once it settled to equilibrium.
Tf = °C

Added by Gregory A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prabhu Ramji

07/06/2022

Step 1

We can use the equation Q_lost = Q_gained to solve for the change in temperature. Show more…

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Ti = initial temperature of the ice cube Tf = final temperature of the puddle m = mass of the ice cube c_ice = specific heat of ice Lf = latent heat of fusion c_water = specific heat of water Q_lost = heat lost by steam Q_gained = heat gained by ice and water Q_lost = mcΔT Q_gained = mc_iceΔT + mL_f + mc_waterΔT Q_lost = Q_gained m_steamc_steamΔT = mc_icec_iceΔT + m_cwaterΔT + mc_waterΔT m_steamc_steamΔT = mc_icec_iceΔT + mc_waterΔT + mc_waterΔT m_steamc_steamΔT = mc_icec_iceΔT + 2mc_waterΔT m_steamc_steamΔT - 2mc_waterΔT = mc_icec_iceΔT ΔT(m_steamc_steam - 2mc_water) = mc_icec_ice ΔT = (mc_icec_ice) / (m_steamc_steam - 2mc_water) ΔT = (0.685kg * 2093J/kg°C * 12.4°C) / (6.79mol * 2009J/(kg°C) - 2 * 0.