Calculating Heat Energy & Temperature Change: Seeking Help!

[Vigo]

1.) An iceberg's volume is 2*10^6 m^3. If the temperature of the iceberg is is 250 K and the density is .9998 g/cm^3, how much heat energy would be required to competely melt it?

2.) An object weighs 500 grams and has a specific heat of 2 Joules/g C. The object absorbs 40% of the output of a 1200 watt dryer. How long will it take to raise the object's temperature from -10 C to 20 C?

Please help me with these problems. My teacher gave them to us for extra credit because he hasn't taught us how to do them. I could really use some extra credit in that class. Thanks.

 

[Andrew Mason]

1.) An iceberg's volume is 2*10^6 m^3. If the temperature of the iceberg is is 250 K and the density is .9998 g/cm^3, how much heat energy would be required to competely melt it?

2.) An object weighs 500 grams and has a specific heat of 2 Joules/g C. The object absorbs 40% of the output of a 1200 watt dryer. How long will it take to raise the object's temperature from -10 C to 20 C?

Please help me with these problems. My teacher gave them to us for extra credit because he hasn't taught us how to do them. I could really use some extra credit in that class. Thanks.

I would suggest that you read ahead in the textbook or the web (eg. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html#c1")on specific heat and "heat of fusion" of water first and then try them.

AM

 

[kyle8921]

Hello there,

Calculating heat energy and temperature change can be tricky, but with some basic principles and equations, we can solve these problems.

For the first question, we need to use the formula Q = mcdeltaT, where Q is the heat energy, m is the mass, c is the specific heat, and deltaT is the change in temperature. We know the density of the iceberg, so we can calculate its mass by multiplying the density by the volume. This gives us a mass of 1.9996*10^9 grams. We also know the temperature change, which is from 250 K to 0 K (since the iceberg will melt completely). Plugging in these values, we get Q = (1.9996*10^9 g)(2 J/g C)(250 K - 0 K) = 9.998*10^11 J. So, it would take approximately 9.998*10^11 Joules of heat energy to completely melt the iceberg.

For the second question, we need to use the formula P = mc(deltaT/t), where P is the power, m is the mass, c is the specific heat, deltaT is the change in temperature, and t is the time. We are given the power output of the dryer (1200 watts), the mass of the object (500 grams), and the specific heat (2 J/g C). We can calculate the change in temperature, which is from -10 C to 20 C, giving us a deltaT of 30 C. We also know that the object absorbs 40% of the output of the dryer, so we can calculate the power absorbed by the object to be (40%)(1200 watts) = 480 watts. Plugging in these values, we get 480 watts = (500 g)(2 J/g C)(30 C/t). Solving for t, we get t = 480 watts * t / (500 g)(2 J/g C)(30 C) = 8 seconds. So, it would take approximately 8 seconds for the object to reach a temperature of 20 C.

I hope this helps you with your extra credit assignment. Remember, it's important to understand the principles and equations behind these calculations, rather than just memorizing the answers. Keep up the good work in your science class!