A metal rod weighing 5.00 g and having a specific heat of 0.397 J / K / g in 22.7 ml of water is placed. The specific heat of the water is 4.
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Question

# <pre class="ql-syntax">A metal rod weighing <span

class="hljs-number">5.00</span> g <span class="hljs-keyword">and</span> having a specific heat <span class="hljs-keyword">of</span> <span class="hljs-number">0.397</span> J / K / g <span class="hljs-keyword">in</span> <span class="hljs-number">22.7</span> ml <span class="hljs-keyword">of</span> water <span class="hljs-keyword">is</span> placed. The specific heat <span class="hljs-keyword">of</span> the water <span class="hljs-keyword">is</span> <span class="hljs-number">4.184</span> J / K / g <span class="hljs-keyword">and</span> the density <span class="hljs-keyword">of</span> the water <span class="hljs-keyword">is</span> approximated to <span class="hljs-number">1000</span> g / mL. The initial water temperature <span class="hljs-keyword">is</span> <span class="hljs-number">20.0</span>oC <span class="hljs-keyword">and</span> the initial temperature <span class="hljs-keyword">of</span> the metal rod <span class="hljs-keyword">is</span> <span class="hljs-number">80.9</span>oC. What <span class="hljs-keyword">is</span> the final temperature (<span class="hljs-keyword">in</span> oC) <span class="hljs-keyword">of</span> the water <span class="hljs-keyword">and</span> the metal rod? </pre>

Solution
(2 metal rod = - &amp;water
M XCXAT = - CmxCXDTD
mass of water 2 22.7 milt wog /ml = 2270 0
5+0. 397 (TF - 80.90) = - (2270* 21-184 (TP -20)
1 + 9 85 ( (f - 80 . () = - 29497-68 ( TF - 20...

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