Question

СРОЧНО До какой температуры можно нагреть 2л воды, взятой при температуре 20°С, если опустить в воду алюминиевый груз массой 200г, нагретый до 150°С? Потерями тепла пренебречь. С РЕШЕНИЕМ, ПОЖАЛУЙСТА

Answer 1

t0=20°C

V1=2л=2*10^-3 м³

ρ1=1000 кг/м³

c1=4200 Дж/кг°С

с2=920 Дж/кг°С

m2=200 г=0,2 кг

Δt2=150°C

t-?

Q1=Q2

Q1=c1m1(t-t0)

m1=V1ρ1 ⇒ Q1=c1V1ρ1(t-t0)

Q2=c2m2Δt2

c1V1ρ1(t-t0)=c2m2Δt2

t-t0=(c2m2Δt2)/(c1V1ρ1)

t=(c2m2Δt2)/(c1V1ρ1)   +t0

t=(920*0.2*150) /(4200*0,002*1000)   +20= (27600)/(8400) +20≈23.3°C

Answer 2

Heating 2L of Water with an Aluminum Load

To calculate the final temperature of the water after adding an aluminum load, we can use the principle of conservation of energy. The heat gained by the water will be equal to the heat lost by the aluminum load.

Given: - Initial temperature of water (T1) = 20°C - Initial temperature of aluminum load (T2) = 150°C - Mass of water = 2 kg (since the density of water is 1 kg/L) - Mass of aluminum load = 200 g = 0.2 kg - Specific heat capacity of water (c1) = 4.18 kJ/kg°C - Specific heat capacity of aluminum (c2) = 0.897 kJ/kg°C

Solution:

The heat gained by the water (Q1) can be calculated using the formula: Q1 = m1 * c1 * (Tf - T1), where: - m1 = mass of water - c1 = specific heat capacity of water - Tf = final temperature of the water

The heat lost by the aluminum load (Q2) can be calculated using the formula: Q2 = m2 * c2 * (T2 - Tf), where: - m2 = mass of aluminum load - c2 = specific heat capacity of aluminum - T2 = initial temperature of aluminum load

Since there are no heat losses, the heat gained by the water is equal to the heat lost by the aluminum load: Q1 = Q2

By substituting the given values and solving for Tf, we can find the final temperature of the water.

Calculation:

Q1 = m1 * c1 * (Tf - T1) Q2 = m2 * c2 * (T2 - Tf)

Setting Q1 equal to Q2: m1 * c1 * (Tf - T1) = m2 * c2 * (T2 - Tf)

Solving for Tf: m1 * c1 * Tf - m1 * c1 * T1 = m2 * c2 * T2 - m2 * c2 * Tf m1 * c1 * Tf + m2 * c2 * Tf = m1 * c1 * T1 + m2 * c2 * T2 Tf * (m1 * c1 + m2 * c2) = m1 * c1 * T1 + m2 * c2 * T2 Tf = (m1 * c1 * T1 + m2 * c2 * T2) / (m1 * c1 + m2 * c2)

Now, we can substitute the given values and calculate Tf.

Final Temperature Calculation:

m1 = 2 kg c1 = 4.18 kJ/kg°C T1 = 20°C m2 = 0.2 kg c2 = 0.897 kJ/kg°C T2 = 150°C

Substituting these values into the formula: Tf = (2 * 4.18 * 20 + 0.2 * 0.897 * 150) / (2 * 4.18 + 0.2 * 0.897)

Calculating Tf gives: Tf ≈ 23.4°C

Answer:

The final temperature of the water, after adding the aluminum load, would be approximately 23.4°C.