0.11cal/g degrees C
A calorie is the amount of heat you need to raise the temperature of one gram of water by one degree Celsius. Assuming you are raising the temperature of the water from twenty degrees Celsius to ninety-nine degrees Celsius, it would take 20,000 calories. To calculate this, subtract 20 from 99. This is the amount of degrees you need to raise the temperature of the water by. Then multiply that number by 256, the amount of water in grams. You should get 20,244 calories. In significant digits, your answer should be 20,000 calories.
The relevant equation behind this problem is Q=m*c* ΔT Where Q is the energy that must be added to or taken from the system, m is the mass of the object, c is the objects specific heat, and ΔT is the change in temperature in Celsius or Kelvin. Plugging in the given values we get that Q=.015kg * 128J/(kg*C) * 10C=19.2J. Therefore, you need 19.2 joules of heat in order to raise the temperature of a .015kg sample of lead by 10 degrees Celsius.
Yes but it can be converted into Celsius
149 degrees Fahrenheit = 65 degrees Celsius.
Use this equation to convert Celsius/Centigrade to degrees Kelvin: [K] = [°C] + 273.15
488.25 J/kg/*C
Every substance has a specific heat. The definition of specific heat is: The amount of energy, usually measured in calories, needed to raise the temperature of one gram of a certain substance by one degree Celsius.
use the equation q=mc∆t, where q is the calories required, m is mass, c is specific heat capacity and ∆t is the change in temperature. Therefore... q=(40g)(0.06 cal/g◦c)(88c - 20c) q=163.2 calories
The amount of heat required to increase the temperature of the substance to 1 degree greater than that of the initial temperature of the body!
80.5 calories 35-12=23 23*3.5=80.5 1c raises 1 gr. h2o 1degree centigrade Here is the formula, it should help a lot:Total Number of Calories = (Specific Heat of Water) ×(Mass of Water) × (Absolute Temperature Change)
Specific temperature is an amount of heat per unit mass required to raise the temperature by one degree Celsius.
The 10 kg of water. This is because the Specific Heat Capacity for water is 9x larger than Iron. The SHC for water is that it takes 1 calorie of water to raise 1 gram by 1 degrees celsius (1C/1gram X 1 degrees celsius) for Iron it takes 0.11 Calories to raise 1 gram by 1 degrees celsius. Therefore meaning since Water has a higher Specific Heat Capacity, it'll take longer to cool down. Just as it requires a large amount of heat to raise the temperature by 1 degrees Celsius, it requires a large amount of heat to drop the temperature by 1 degrres Celsius. - Ataa Ghomashchi
For one gram of ice, it takes 11.9 calories to change the temperature to 0°, 80 calories to melt the ice, 100 calories to raise the water temperature to 100°, 540 calories to change the water to steam, and 23 calories to raise the steam temperature to 123°. That's a total of (11.9 + 80 + 100 + 540 + 23) calories or 754.9 calories. So to do the same to 55.6 grams of ice requires 55.6 times as much heat. 754.9 calories times 55.6 equals approximately 41972 calories (about 42 kilocalories).
specific gravity of gasoline at 104 celsius
2.460 is the specific heat of ethyl alcohol in J/gC, so I will find joules and then convert to calories. q(joules) = mass*specific heat*change in temp. q = (13 g alcohol )(2.460 J/gC)(23 C - 11 C) = 383.76 joules (1 calorie/4.184 joules) = 92 calories ------------------
We know that Q=m.s.t, where Q= Heat, s= Specific heat of the substance, t=temperature(Difference in temperatures) =>s= Q/m.t =>s=525/(25X15) =>s=525/375 =>s=1.4 cal/g/0c Specific heat of the substance is 1.4 cal/g/0c.
quantity of heat required =mass*specific heat of water*change in temperature Q=160*1*(77-19)=9280 calories=9.28 kilo calories