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How fast the energy is provided (power, in joules/second or watts) is irrelevant, as long as not too much energy gets radiated away. What you really need to know is how much energy (in joules) is needed.
you need to know the specific heat capacity of aluminum first which can be denoted as 'c'. then using the formula: H=MC(dt) where 'dt; is the change in temperature, M is the mass and H is energy needed, you can thus calculate H. i think the value of 'c' for aluminum is around 0.88Jg/K Replace given values in the equation, you will surely get the answer!
It takes 4.186 Joules to heat one gram of water by 1-degree Celsius. 4.186 * 4000 = 16,744 Joules to heat 4 kilos of water by 1-degree. 16,744 * 70 = 1,172,080 Joules. The above assumes that one litre of water weighs exactly 1 Kilogram.
Approx 2940 Joules.
It requires 1 calorie to increase 1 mL of water by 1 degree Celsius. In 68 grams of water, there are 68 mL. The change in temperature is 7 degrees, so 476 calories are needed. 1 calorie has 4.184 joules, hence 1.992 kJ are needed.
1 kcal equal 1 000 calories or 4,184 joules. 1 calorie is equal to the heat needed to increase the temperature of 1 g water with 1 0C.
How fast the energy is provided (power, in joules/second or watts) is irrelevant, as long as not too much energy gets radiated away. What you really need to know is how much energy (in joules) is needed.
134 joules. You're very welcome for answering your question.
you need to know the specific heat capacity of aluminum first which can be denoted as 'c'. then using the formula: H=MC(dt) where 'dt; is the change in temperature, M is the mass and H is energy needed, you can thus calculate H. i think the value of 'c' for aluminum is around 0.88Jg/K Replace given values in the equation, you will surely get the answer!
51,520 Joules must be added to increase it's temperature to 100 C.
8.200 J
E = mass x specific heat x Δ°T Δ°T = new temperature - original temperature where Δ°T is equal to temperature change (Celsius in this case). The specific heat of Al is 0.900 J/g°C. Before we proceed to find the quantity of heat in joules, we must first find the temperature change. To calculate the temperature change, we must subtract the original temperature from the new temperature. Δ°T = 50°C - 25°C = 25°C In order to find the quantity of heat (joules), we must multiply mass, specific heat, and the temperature change (calculated above). E = 40.0g x 0.900 J/g°C x 25°C = 900 Joules or 9.0 x 102 Joules
It takes 4.186 Joules to heat one gram of water by 1-degree Celsius. 4.186 * 4000 = 16,744 Joules to heat 4 kilos of water by 1-degree. 16,744 * 70 = 1,172,080 Joules. The above assumes that one litre of water weighs exactly 1 Kilogram.
q(Joules) = mass * specific heat * change in temperature q = 32.0 grams H2O * 4.180 J/gC *(54.0 C - 12.0 C) = 5617.92 Joules this is, of course 5.62 kilojoules
Work done (joules) and time taken (seconds) is the information needed to calculate power in watts (joules/second).
Approx 2940 Joules.
The needed heat is 47,65 Joules.