Question

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To find cx, the specific heat of material X, I place 75g of it in a 30g copper calorimeter that contains 65g of water, all initially at 20o C. When I add 100g of water at 80o C, the final temperature is 49o C. What is cx?​

Answer 1

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Answer 2

Given:

The specific heat of material X

place 75g of it in a 30g copper calorimeter that contains 65g of water, all initially at 20° C

I add 100g of water at 80° C

Find:

cx = ?

Solution:

$\left(m_{x} c_{x}+m_{\text {copper }} \mathcal{C}_{\text {copper }}+m_{\text {water }} \mathcal{C}_{\text {water }}\right)\Delta t_{\text {initial.components }}\\=m_{\text {hot.water }} c_{\text {hot.water }} \Delta t_{\text {hot.water }}$

$\text { where both temperature changes are positive numbers. }$

$c_{x}=\frac{1}{m_{x}}\left(m_{\text {hot.water }} c_{\text {hot.water }} \frac{\Delta t_{\text {hot.water }}}{\Delta t_{\text {initial components }}}-m_{\text {copper }} c_{\text {copper }}-m_{\text {water }} c_{\text {water }}\right)$

$\text { If we look up values for the specific heats of water and copper, we can now solve this }$$\text { equation. }$

$c_{x}=\frac{1}{.075 \mathrm{~kg}}\left[(.1 \mathrm{~kg})\left(4184 \frac{J}{\mathrm{~kg} \cdot K}\right)\left(\frac{31 K}{29 K}\right)-(.03 \mathrm{~kg})\left(386 \frac{J}{\mathrm{~kg} \cdot K}\right)-(.065 \mathrm{~kg})\left(4184 \frac{J}{\mathrm{~kg} \cdot K}\right)\right]$$c_{x} \approx 2180 \frac{J}{k g \cdot K}$

Hence the value of $c_{x} \approx 2180 \frac{J}{k g \cdot K}$