A calculate the total amount of energy cal that is needed to...

a) Calculate the total amount of energy (cal) that is needed to melt 10.0 kg of ice at 0.0°C. b) Calculate the total amount of energy (J) that is needed to melt 10.0 kg of ice at 0.0°C c) Calculate the total amount of energy (cal) that is needed to raise the temperature of 10.0 kg of water from 0.0°C to 20.0°C. d) Calculate the total amount of energy (cal) that is needed to melt 10.0 kg of ice at 0.0°C and raise the temperature of the melted water to 20.0°C. e) Calculate the total amount of energy (cal) that is needed to melt 10.0 kg of ice at 0.0°C, raise the temperature of the melted water to 20.0°C, and evaporate half of it.

Transcript

00:01 Here the first question is to calculate the total amount of energy that is q required to convert 10 kilogram of ice to water.

00:10 So the exact conversion can be written as ice at zero degrees celsius is converted to water at zero degrees celsius.

00:17 Now the heat change q is equal to m into f where m is the mass and f is the heat of fusion.

00:23 Now the heat of fusion value in terms of calorie is equal to 80 calorie per gram.

00:28 So we have mass is 10 kilogram multiplied by 80 calorie per gram.

00:36 Now this is in gram.

00:38 So we have to convert this kilogram into gram.

00:40 10 kilogram is equal to 10 into 10 to the power 3 gram, which is equal to 10 into 10 to the power 3 gram into 80 calorie.

00:52 So this is equal to 800 multiplied by 10 to the power 3 calorie, which is the answer to this.

00:58 Question so the next question is to calculate the same in terms of jule so what we have is q is again equal to m into f where f is the heat of fusion now this is equal to 336 into 10 to 3 jule per kilogram so we have 10 kilogram multiplied by 3 36 into 10 to the power 3 jule per kilogram which is equal to 336 into 10 to the power 4 jule.

01:33 So this is the answer to this question.

01:36 The next question is to calculate the amount of energy q that is required to convert 10 kilogram of water at 0 degrees celsius to 10 kilogram of water at 20 degrees celsius.

01:51 And q must be calculated in terms of calorie.

01:53 Now the equation for q for this conversion is mc delta t, where m is the mass c is c is the mass c the specific heat capacity of water and delta t is the change in temperature now specific heat of capacity c in terms of calorie is one calorie per gram per degree celsius and delta t is equal to the final temperature minus the initial temperature is 20 degrees which is minus initial temperature is 0 degrees celsius which is equal to 20 degrees celsius and since the c is in terms of gram so we must convert mass in terms of grams so 10 kilogram is equal to 10 into 10 to the power 3 gram so on substituting this we have mass is 10 into 10 to the power 3 gram into c is 1 calorie per gram per degree celsius multiplied by delta t is 20 degrees celsius this is equal 200 into 10 to the power 3 calorie now the next question b is to convert 10 kilogram of ice at 0 degrees celsius to water at 20 degrees celsius.

03:07 So if we write the steps, ice at 0 degrees celsius is converted to water at 0 degrees celsius which is further converted to water at 20 degrees celsius.

03:19 So we already found this as 800 into 10 to the power 3 calorie.

03:26 And this is 200 into 10 to the power 3 calorie...

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