Question

До якої температури нагріється 5кг води за рахунок 42 кДж енергії, якщо її початкова температура 30'С?​

Answer 1

Ответ: 32°C

Объяснение:

Q=c*m*Δt

Δt=Q/(c*m)=42000/(4200*5)=2°

t=to+Δt=30+2=32°C

Answer 2

Calculation of the Final Temperature of Water

To determine the final temperature of water after absorbing a certain amount of energy, we can use the equation:

Q = mcΔT

Where: - Q is the energy absorbed by the water (in joules) - m is the mass of the water (in kilograms) - c is the specific heat capacity of water (approximately 4.18 J/g°C or 4186 J/kg°C) - ΔT is the change in temperature of the water (in °C)

In this case, we have: - Q = 42 kJ = 42,000 J - m = 5 kg - c = 4186 J/kg°C - Initial temperature (T1) = 30°C

We need to find the final temperature (T2).

Using the equation above, we can rearrange it to solve for ΔT:

ΔT = Q / (mc)

Substituting the given values:

ΔT = 42,000 J / (5 kg * 4186 J/kg°C)

Calculating the value of ΔT:

ΔT ≈ 2.52°C

To find the final temperature (T2), we add the change in temperature (ΔT) to the initial temperature (T1):

T2 = T1 + ΔT

Substituting the values:

T2 = 30°C + 2.52°C

Calculating the value of T2:

T2 ≈ 32.52°C

Therefore, the final temperature of 5 kg of water, after absorbing 42 kJ of energy, starting from an initial temperature of 30°C, is approximately 32.52°C.