Question

В калориметре находится вода массой 0,4 кг при температуре, равной 10 градусов, и нормальном атмосферном давлении. В воду положили кусок льда массой 0,6 кг при температуре, равной -40. Какая температура установится в калориметре? Теплоемкостью материала из которого изготовлен калориметр, и тепловыми потерями пренебречь.

Answer 1

Ответ:

Q₁=c₁m₁*10=4200*0.4*10=16800 Дж - кол-во теплоты для охлаждения воды до 0

Q₂=λm₁=334000*0.4=133 600 Дж - кол-во теплоты для замерзания воды

Q₃=c₂m₂*40=2100*0.6*40=50 400 - кол-во теплоты для нагревания льда до 0

Q₄=λm₂=334000*0.6=200 400 Дж - кол-во теплоты для плавления льда

Q₃<Q₁+Q₂

Q₄>Q₁+Q₂-Q₃

Значит температура будет равна 0°С

Answer 2

Calculation of Final Temperature in the Calorimeter

To calculate the final temperature in the calorimeter after adding a piece of ice, we can use the principle of conservation of energy. The heat gained by the ice will be equal to the heat lost by the water in the calorimeter.

Let's denote the specific heat capacity of water as Cw and the specific heat capacity of ice as Ci. The specific heat capacity represents the amount of heat required to raise the temperature of a substance by 1 degree Celsius per unit mass.

Given: - Mass of water (mw) = 0.4 kg - Initial temperature of water (Tw) = 10 degrees Celsius - Mass of ice (mi) = 0.6 kg - Initial temperature of ice (Ti) = -40 degrees Celsius

We can use the following equation to calculate the final temperature (Tf):

mw * Cw * (Tf - Tw) = mi * Ci * (Tf - Ti)

Since we are asked to neglect the heat capacity of the calorimeter material and any heat losses, we can assume that the heat gained by the ice is equal to the heat lost by the water.

Now, let's substitute the given values into the equation and solve for Tf.

0.4 * Cw * (Tf - 10) = 0.6 * Ci * (Tf - (-40))

Unfortunately, the specific heat capacities of water and ice were not provided in the search results. Therefore, we cannot provide an exact numerical answer without this information. However, you can use the equation above and substitute the specific heat capacities of water and ice to calculate the final temperature in the calorimeter.