It takes 1 calorie of heat to raise 1 gram or water or ice 1° C, but it takes 80 calories to change 1 gram of ice at 0° C to 1 gram of water at 0° C. It's best to divide the problem into three parts, raising the temperature of the ice, melting the ice, and raising the temperature of the water, then combine the three parts at the end. You can just as easily do it in two parts, combining the two parts in which the temperature is raised, but I'll do it the less confusing way this time.
ice @ -20° to ice @ 0°:
30 g * 20° * 1 cal/°/g = 600 cal
ice @ 0° to water @ 0°:
30 g * 80 cal/g = 2400 cal
water @ 0° to water @ 20°:
30 g * 20° * 1 cal/°/g = 600 cal
(600 + 2400 + 600) cal = 3600 cal
(or 3.6 kilocalories)
The idea here is to: * Look up the specific heat of water. * Multiply the mass, times the temperature difference, times the specific heat of water. You may need to do some unit conversions first; specifically, if the specific heat is given per kilogram, you can convert the grams to kilograms.
Raise the temp of 52 grams of water from 33.0 C to 100 C = 52*67*4.184 = 14.577 kJConvert evaporate 52 g of water to steam without change of temp = 52*2259 = 117.468 kJRaise the temp of 52 grams of steam from 100 C to 110 C = 52*10*2.02 = 1.051 kJTotal energy required = 133.095 kJ = 31,811 calories or 31.811 kCal.
Density = grams/ml 1.00 g/ml = X g/5.0 ml = 5.0 grams water ============== q(joules) = mass * specific heat * change in temp. q = (5.0 grams)(4.180 J/gC)(75 C - 2.50 C) = 1515.25 Joules ---------------------------------/4.184 = 362 calories -------------------
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.
The density of sulfur at 25 degrees Celsius is approximately 2 grams per cubic centimeter.
The idea here is to: * Look up the specific heat of water. * Multiply the mass, times the temperature difference, times the specific heat of water. You may need to do some unit conversions first; specifically, if the specific heat is given per kilogram, you can convert the grams to kilograms.
The solubility of ammonium chloride at 60 degrees Celsius is approximately 39.2 grams per 100 grams of water.
The solubility of potassium chloride at 25 degrees Celsius is approximately 34 grams per 100 grams of water.
To convert 12.5 grams of ice at 0 degrees Celsius to steam at 100 degrees Celsius, you would need to provide heat energy for three main processes: heating the ice from 0 degrees Celsius to 100 degrees Celsius, melting the ice into water at 0 degrees Celsius, and then heating the water from 0 degrees Celsius to steam at 100 degrees Celsius. The total calorie requirement would be determined by the specific heat capacities and heat of fusion and vaporization of water.
The density of helium gas at 25 degrees Celsius is approximately 0.1785 grams per liter.
Approximately 39 grams of sodium chloride can be dissolved in 100 grams of water at 95 degrees Celsius.
The density of 1-propanol at 25 degrees Celsius is approximately 0.804 g/cm3.
Density of ice at 0 degrees Celsius is 916.8 grams per cubic centimeter or milliliter. The density of fresh water is dependant on the temperature: At 3.98 degrees Celsius the density is 0.999975 grams per milliliter. At 100 degrees Celsius the density is 0.958.35 grams per milliliter.
Approx 4974 Joules.
The density of oxygen at 25 degrees Celsius is approximately 1.1839 grams per liter (g/L).
To decrease the freezing point of water by 2.5 degrees Celsius, you would need to dissolve approximately 37.5 grams of sugar in 300 grams of water. This is calculated based on the colligative property that states freezing point depression is directly proportional to the molality of the solute in the solution.