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lyudmila [28]
3 years ago
15

A 60.0 g sample of chromium at 82.0°C (specific heat for chromium is 0.11 J/g°C) was placed in 80.0 g of water. Assume there is

no loss of heat to the environment. What is the temperature of the water and the chromium? The water's initial temperature was 24.0°C.
Chemistry
1 answer:
jeka943 years ago
3 0

The temperature of the water and the chromium : 25.122 ° C  

<h3>Further explanation  </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released  

Q in(gained) = Q out (lost)

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q Chromium= Q water

\tt 60\times 0.11\times (82-t)=80\times 4.18\times (t-24)\\\\541.2-6.6t=334.4t-8025.6\\\\341t=8566.8\\\\t=25.122^oC

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<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

We are given:

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Mass percentage of C_2H_2O = 15 %

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We know that:

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To calculate the average molecular mass of the mixture, we use the equation:

\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}

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\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}

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<u />

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