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Typically, water runs through the baseboard copper tubing and, therefore, fresh hot water is constantly running through the piping. However, consider a pipe where water was allowed to sit in the pipe. The hot water cools as it sits in the pipe. What is the temprature change, (ΔT), of the water if 194.0 g of water sat in the copper pipe from part A, releasing 1238 J of energy to the pipe? The specific heat of water is 4.184 (J/g)⋅∘C.

Sagot :

sebassandin

Answer:

[tex]\Delta T=1.53\°C[/tex]

Explanation:

Hello!

In this case, since the problem is providing mass, energy, and specific heat of water, we can write the following equation that relates them to each other:

[tex]Q=mC\Delta T[/tex]

In such a way, we get rid of both m and C, to solve for the change in temperature:

[tex]\Delta T=\frac{Q}{mC}[/tex]

Now we plug in the data to obtain:

[tex]\Delta T=\frac{1238J}{194.0g*4.184\frac{J}{g\°C} }\\\\\Delta T=1.53\°C[/tex]

Best regards!

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