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3 years ago

If 2,035 cal of heat is added to a 500.0 g sample of water at 35.0°C, what is the final

Chemistry

1 answer:

Lynna [10]3 years ago

5 0

Answer:

39.1 °C

Explanation:

Recall the equation for specific heat:

$q=mc \Delta T\\$

Where q is the heat, m is the mass, c is the specific heat of the substance (in this case water), and delta T is the change in temperature.

You should know that the specific heat of water is 1 cal/g/C.

Using the information in the question:

$2035=500(1)(T-35)\\2035=500T-17500\\500T=19535\\T=39.07$

The final temperature is about 39.1 °C.

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3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,

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Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

$Q_{Ag}=-Q_{liquid}$

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$m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})$

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

$Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}$

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