Question

9:484 Back Specific Heat Chpt3.docx Heat Capacity 1) How many Joules of energy are required to raise 15g of aluminum from 30 C to 100°C? Given the heat capacity of Aluminum is 0.903 Jig C 2) If 6093 cal of energy is added to some pure substance. If the temperature went from 10°C to 50°C, how many grams of the substance was present. Given the of the substance is 2.45cal/g C 3) If 4x10* J of energy is added to 43g of a pure substance ant the temperature rose 68°C, what is the heat capacity of the pure substance? 4) a) 45 ml of water at 25°C has 202 cal. of energy added to it. What will be the temperature change the water undergoes? (45ml water = 45g), ČIcalig"C b) What will be the final temperature the water is problem 4a reaches 5) If 403 J of energy is added to 8.1g of iron at 20°C, what is the maximum temperature the iron will reach? Given 0.4491g C Key: 1)948) 2)6283) 13.71gC4) a) 4.5°C b) 29.3°C 5) 131°C 35 D G] Dashboard Calendar To Do Notifications Inbox

Answer 1

Given mass of Aluminium (m) = 15 g change in temperature (OT) = 100°C-30°C - 70°C Heat capacity (s) = 0.903J/gcc To find Energy required in Joules COH We know that, OH = mSAT =(159) (0.903 J/gº) (70°C) =948.15 J Thus, energy required is 948. 15 J

2:- Giren: Energy сон) added 6093 cal change in temperature COT) = 50 -10°= 40°C Heat capacity (cle) = 7.45 callgºc To find mass of substance (m)

We know that, OH = mSAT 6093 cal = mx(7.45 callgº) x64099 m = 6093 62.17 g 2.45X40 Thus, mass of substance = 62.17g.

@3:- Cah) Given: Energy supplied = 4x10u J mass of substance (m) = 439 change in temperature COT)= 68°C To find :- heat capacity (5) We know that, OH= mSAT -> 4x1045 = (439)x5 x (68°9 - s = uxlou 13.67 J/g °C 43x68 Thus, Heat capacity of substance is 13.67J/goc

Giren : mass of water (m) = 45g Energy added (OH)= 200 cal Heat capacity (c) = 1 callgie To find change in temperature. Assuming, water do not change its state and all the energy is used to raise the tempera- -ture only. Therefore, Using, AH = MCAT => 202 cal = (459) (1 callgº) (OT) = AT= - 4.49°C 202 us

(6 Therefore, final temperature = initial temperature tot = 25°ct 4.49°C 29.49°C Thus, final temperature of water is 29.49°C

5) Giren : Energy added (OH) = 403J mass of iron (m) = 8.1g a Heat capacity (s) = 0.44gJ/g °C Initial temperature = 20°C To find: We know Final temperature. that, OH = m SDT => 403J = (8.lg) (0.4495/7%)%07 => AT= 403 8.1*0.44g 110.808°C Therefore, final temperature initial temp. + at = 20°С+ 110.808°C ř 130.81°C Thus, maximum temperature of iron is 130.81°C
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