What amount of heat will give a glass of boiling water with a volume of 750 cm ^ 3
What amount of heat will give a glass of boiling water with a volume of 750 cm ^ 3, cooling down to a temperature of 28 grams?
The formula that we can apply in this case is Q = cm (t start – t end), where Q is the amount of heat released during the cooling of the body, where c is the specific heat capacity of the substance, m is the body weight, t start is the temperature, to which the body was first heated, t
final – the final temperature to which the body has cooled down.
(t start – t end) is the temperature difference, that is, its value, by which it has cooled
body, in our case – from 100 to 28 ° С (72 ° С).
The specific heat capacity of water is 4183 J / kg ° С.
Let’s assume that the volume is 750 cm³ of water, which means a mass of 0.75 kg (the density of water is 1000 kg / m³ or 1 kg / l (1 l = 1000 cm³)).
Substituting the values into the above formula, we get – Q = (4183 J / kg ° C) * (0.75 kg) * (100 ° C – 28 ° C).
Answer: when a glass of water cools down according to the given conditions of the problem, an amount of heat will be released, approximately equal to – Q = 225.882 kJ.
