Question

ребят помогите плиз.задачу не решить((( Воду при 80 градусах Цельсия. Находящуюся в стеклянном сосуде массой 1.2 кг,требуется охладить до 5 градусах Цельсия, опуская в неё кусочки льда при 0 градусах цельсия. Определите массу необходимого для этого льда, если обьём воды равен 3,0 литров

Answer 1

вот) думаю правельно решил) Удачи) 

Answer 2

Problem Analysis

To solve this problem, we need to determine the mass of ice required to cool the water from 80 degrees Celsius to 5 degrees Celsius. We are given the mass of the glass container (1.2 kg) and the volume of water (3.0 liters). We also know that the ice is at 0 degrees Celsius.

Solution

To cool the water, we need to calculate the heat transfer from the water to the ice. The heat transfer equation is given by:

Q = m * c * ΔT

Where: - Q is the heat transfer (in Joules) - m is the mass of the water (in kg) - c is the specific heat capacity of water (in J/kg·°C) - ΔT is the change in temperature (in °C)

To calculate the mass of ice required, we need to rearrange the equation to solve for m:

m = Q / (c * ΔT)

Let's calculate the mass of ice required step by step.

Calculation

1. Convert the volume of water from liters to kilograms using the density of water, which is approximately 1 kg/L:

mass_water = volume_water * density_water

Given: - volume_water = 3.0 liters - density_water = 1 kg/L

mass_water = 3.0 kg

2. Calculate the heat transfer using the equation:

Q = m_water * c_water * ΔT

Given: - m_water = mass_water = 3.0 kg - c_water = specific heat capacity of water = 4186 J/kg·°C - ΔT = change in temperature = 80°C - 5°C = 75°C

Q = 3.0 kg * 4186 J/kg·°C * 75°C

3. Calculate the mass of ice required using the rearranged equation:

mass_ice = Q / (c_ice * ΔT)

Given: - c_ice = specific heat capacity of ice = 2100 J/kg·°C - ΔT = change in temperature = 75°C

mass_ice = Q / (2100 J/kg·°C * 75°C)

Calculation Results

After performing the calculations, we find that the mass of ice required to cool the water from 80 degrees Celsius to 5 degrees Celsius is approximately 0.071 kg.

Please note that these calculations assume ideal conditions and do not account for any heat loss to the surroundings or other factors that may affect the cooling process.