What is the number of calories given off by 500 grams of water cooling from 50 degrees Celsius to 20 degrees Celsius?

Answer 1

The heat of fusion (aka enthalpy of fusion) of ice at 0 °C is about 79.72 cal/g.

The specific heat of water in the range from 30 °C to 0 °C is about 1 cal/g/°C.

The total energy that must be removed is thus:

20.0 g x[(0-30)°C x 1 cal/g/°C - 79.72 cal/g] = -2184 cal

in other words you must remove 2184 cal to take 20.0 g of water from liquid at 30 30 °C to solid (ice) at 0 °C.

Answer 2

I will work in joules then convert.

q = mass * specific heat * change in temperature

q = (500 g H2O)(4.180 J/g*oC)(50o C - 20o C)

= 62700 Joules (1 calorie/4.184 Joules)

= 1.5 X 104 calories

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Answer 3

The specific heat capacity of water is 4.18 J/g°C. To calculate the heat released during cooling, first, find the temperature change (50°C - 20°C = 30°C). Then, use the formula Q = mcΔT (Q = heat energy, m = mass, c = specific heat capacity, ΔT = temperature change) to find the energy released, which is approximately 62700 J or 62.7 kJ. To convert to calories, divide by 4.18 J/g°C = 15000 calories or 15 kcal.

Answer 4

You need to multiply together all of the following:* The mass in grams

* The temperature difference

* The specific heat of water

Answer 5

16,000

=200 (100-80)

Answer 6

200 calories