2 A researcher studying the nutritional value of a new candy places a 5.30-gram

Question

2 A researcher studying the nutritional value of a new candy places a 5.30-gram sample of the candy inside a bomb calorimeter and combusts it is excess oxygen. The observed temperature increase is 2.24 degree C. If the heat capacity of the calorimeter is 43.60 kJ-K^-1, how many nutritional Calories are there per gram of the candy? If you combine 300.0 mL of water at 25.00 degree C and 100.0 mL of water at 95.00 degree C, what is the final temperature of the mixture? Use 1.00 g/mL as density of water.

Explanation / Answer

a) Since - Q = C (DT) = (43.60 kJ/K)(2.24 K) = 94.664 kJ

This heat is generated by 5.30 g of the sample, hence, heat for 1 g

          = (94.664*103 J *1 cal )/ (4.184 J*5.30 g) = 4268.91 J/g

b) Since Qabsorbed = Qevolved

(m)(s)(DT)lower T = (m)(s)(DT)high T

Hence - (300 g)(4.184 J/g.C)(T-25) C = (100 g)(4.184 J/g.C)(95-T)

By solving - 3(T-25) = (95-T)

3 T - 75 = 95 - T

4 T = 170

T = 42.5 C (Final temperature)

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