Heat required = (mass) x (specific heat of substance) x (temp difference)
In this case it would be (100) x (1) x (50-20)
= 100 x 30 = 3000 cals
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).
q (amt of heat) = mass * specific heat * temp. differenceThe specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oCq = (105 grams)*(1.00 cal/goC)*(40oC)= 4,200 calories
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 -------------------
the less force is needed.
The specific heat of air at 0 degrees Celsius is 1.01 Joules per gram or J/g. The specific heat of a substance is defined as the quantity of heat per unit mass needed to raise its temperature by one degree Celsius.
100 calories. 1 calorie is defined as the amount of energy required to raise the temperature of 1 gram of 1 by 1 degree Celsius. So, if you need to raise 10 grams of water 1 degree, you would need 10 calories of energy. If you needed to raise those same 10 gram by 10 degrees, you'll need 10 * 10, or 100 calories.
The answer is 2 calories.
1
(5)(3)= 15 calories. 1 calorie is the energy (heat) to raise 1 gram of water by 1 degree celsius, so 5 grams of water (3 degrees Celsius) = 15.
1 calorie is defined as the amount of energy needed to raise the temperature of 1 gram of water by 1C, so... It takes 8.1 calories to raise your 8.1 grams by 1C, but you need to raise it 20C. 8.1*20=162. 162 calories is the answer you are looking for.
To raise the temperature of 1 gram of ice by 1 degree Celsius, it requires 0.5 calories. Therefore, to raise the temperature of 60 grams of ice by 1 degree Celsius, it would require 30 calories.
There are two different kinds of calories, cal or Cal. Einstein uses calories (cal) as the energy needed to raise the temperature of 1 gram of water through 1 degree Celsius. The large calorie (Cal) is the energy needed to raise the temperature of 1 kilogram of water through 1 degree Celsius. This large calorie (Cal) is equal to one thousand small calories and often used to measure the energy value of foods.
It takes 2.44 calories of heat to raise the temperature of one gram of ethyl alcohol by 1 degree Celsius.
The amount of heat needed to raise the temperature of a substance is given by the formula: Q = mcΔT, where Q is the heat (in calories), m is the mass (in grams), c is the specific heat capacity of water (1 cal/g°C), and ΔT is the change in temperature. Plugging in the values: Q = 8g * 1 cal/g°C * 7°C = 56 calories.
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.
Yes, the amount of energy that food contains is measured in calories. When we refer to calories in the context of food, we are actually referring to kilocalories (kcal), which represent the amount of energy needed to raise the temperature of 1 kilogram of water by 1 degree Celsius.
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).