Question
To cool a very hot piece of $4.00-\mathrm{kg}$ steel at $900^{\circ} \mathrm{C}$, the steel is put into $0.5 .00-\mathrm{kg}$ water bath at $20^{\circ} \mathrm{C} .$ What is the final temperature of the steel-water mixture?
Step 1
The formula for this is $Q_1 = m_w c_w \Delta T_w$, where $Q_1$ is the heat gained by water, $m_w$ is the mass of water, $c_w$ is the specific heat of water, and $\Delta T_w$ is the change in temperature. Show more…
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