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Suppose you want to heat 40 g of water by 20 degrees celsius how many joules of heat are required?
The amount of heat required can be calculated using the formula: Q = mc∆T, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ∆T is the change in temperature. Plugging in the values (m = 40g, c = 4.18 J/g°C, ∆T = 20°C), you would find the amount of heat required in joules.
As the lake surface freezes in the winter how many joules of heat are realized by each gram of water?
When water freezes into ice, it releases approximately 334 joules of heat per gram. This process is known as the latent heat of fusion, which is the energy required to change water from a liquid to a solid without changing its temperature. Therefore, as the lake surface freezes in winter, each gram of water realizes 334 joules of heat.
As the lake surface freezes in the winter how many joules of heat are released by each gram of water?
When water freezes into ice, it releases approximately 334 joules of heat per gram. This process is known as the latent heat of fusion, which is the energy required to change water from a liquid to a solid without changing its temperature. Thus, as the lake surface freezes, each gram of water releases about 334 joules of heat into the surrounding environment.
How many joules of energy are necessary to heat a smaple of water with a mass of 46.0 grams from 0.0 c to 100.0 c?
Water has a specific heat of 4.18J/gC, so set up the specific heat equation: C (spec. heat)=q (joules)/mass x temp. change, so: 4.18 (spec. heat of water) = q/46g(100deg), so q = 4.18(4,600) = 19,228 joules (or 19.228 kJ).
What is the standard unit of heat?
The standard unit of heat is the calorie. It is defined as the amount of heat required to raise the temperature of 1 gram of water by 1 degree Celsius. Another commonly used unit is the joule, with 1 calorie equal to 4.184 joules.
Suppose you want to heat 40g of water by 20 degrees Celsius how many joules of heat are required?
Use the equation q=mc(delta t) (that is, heat equals mass times specific heat times the change in temperature) to answer the question. The specific heat of water is 4.186 Joules per gram-Celsius. Therefore, q=(40)(4.186)(20), which equals 3348.8 Joules of heat (or approximately 3.35 kiloJoules of heat).
Suppose you want to heat 40kg of water by 20c How many joules of heat are required?
Energy = mass x specific heat capacity x change in temperature specific heat capacity of water = 4200 Joule per kilogram per degree Celsius Energy = 40 x 4200 x 20 = 3.36x106 Joules.
Suppose you want to heat 40 g of water by 20 degrees celsius how many joules of heat are required?
The amount of heat required can be calculated using the formula: Q = mc∆T, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ∆T is the change in temperature. Plugging in the values (m = 40g, c = 4.18 J/g°C, ∆T = 20°C), you would find the amount of heat required in joules.
How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams for 0.0?
To calculate the energy required to heat water, you would need to know the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g°C. Assuming we are heating the water by 1°C, the energy required would be 46.0g * 4.18J/g°C * 1°C = 192.28 Joules.
What is the heat of vaporization of water in joules per kilogram?
The heat of vaporization of water is 2260 joules per kilogram.
How much joules does it take to heat water to 100 degrees?
To heat 1 gram of water by 1 degree Celsius, it takes 4.18 joules. So, to heat water from, for example, 20 degrees to 100 degrees, you would need to calculate the total mass of water and apply the specific heat capacity to determine the total energy required.
What is the latent heat of vaporization of water in joules per kilogram?
The latent heat of vaporization of water is 2260 joules per kilogram.
How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 celsius to 100.0?
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6
How much energy is required to raise water temp 1 degree Celsius?
The amount of energy required to raise the temperature of water by 1 degree Celsius is known as its specific heat capacity. For water, the specific heat capacity is 4.18 Joules/gram°C. This means that it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
How many joules of energy are necessary to heat a smaple of water with a mass of 46.0 grams from 0.0 c to 100.0 c?
Water has a specific heat of 4.18J/gC, so set up the specific heat equation: C (spec. heat)=q (joules)/mass x temp. change, so: 4.18 (spec. heat of water) = q/46g(100deg), so q = 4.18(4,600) = 19,228 joules (or 19.228 kJ).
How many joules are required to boil 21.1 g of water at 100 C?
The heat required to boil water can be calculated by multiplying the mass of water (21.1 g) by the specific heat capacity of water (4.18 J/g°C) and the temperature change (100°C - initial temperature). This calculation results in 8.82 kJ or 8820 J of energy needed to boil 21.1 g of water at 100°C.
What is the specific heat of water in joules per kilogram degree Celsius?
The specific heat of water is 4186 joules per kilogram degree Celsius.
