Tag Archives: specific heat problems examples

Specific Heat Problems Examples

specific heat problems examples
Should I use this value for the specific heat in the equation?

In my chemistry class, we work with the equations of heat. The equation we use is heat = specific heat (calories / gram x 1 degree) Celsius temperature change x mass x (delta t). Let me give an example of one of the homework Problems: A sample of 200 g water at 90 in Degrees Celsius is heated to boiling 100th Degrees Celsius, and then converted into steam at 100. Degrees Celsius. How many calories are needed the sample heat? My answer was 110.00 calories. The heat of vaporization of water is 540 c / g. My question is: am I the heat of evaporation rather than the specific Heat of the equation?

You asked if you must use one or the other, but you must use both the specific heat and heat of vaporization This issue has to first heat the water 90-100 1 cal / gC ° * C * 10 = 2000 cal 200 g and then we need to put energy cause phase change, now that You have a device, add the boiling point of 200 g * 540 cal / g = 108.000 cal now those two together and get 110 000 calories. I got the same answer as you, so we probably she is right. If you happen to ice cold to hot steam you have the energy to heat the ice to zero, then to melt the heat of fusion, then the specific heat again to get it to 100 heat, then the heat energy of the evaporation of the boil, then further specific heat energy, the temperature of the vapor (specific Increase heat is not really the same at all temperatures, but liquid water, it is not change much between 0 and 100, but the value of ice and steam very is different)

Specific Heat Problem Part 1 (Student Example)

Specific Heat Problems

specific heat problems
Help please! Specific problems of the heat?

No idea what I do. Please help! 1) A sample of aluminum absorbs 300.0 J energy. The mass of the sample was 424 g. What was the change in temperature? CP = 0.899 aluminum

Q = mass x specific heat x delta T = 0.899 x 300 424 x delta T delta T = 0.787 ° C

Specific Heat Word Problem