a banana with a temperature of 17 degrees C (specific heat of banana= 3520j/kgK) and a a mass of 250g is placed in a bowl of milk with a temperature of 8 degrees C and a mass of 350g (specific heat of milk= 4200 J/kgK). no heat is lost to the bowl or the air as the banana and milk reach equilibrium. what is the equilibrium temperature of the banana and milk?

Question

a banana with a temperature of 17 degrees C (specific heat of banana= 3520j/kgK) and a a mass of 250g is placed in a bowl of milk with a temperature of 8 degrees C and a mass of 350g (specific heat of milk= 4200 J/kgK). no heat is lost to the bowl or the air as the banana and milk reach equilibrium. what is the equilibrium temperature of the banana and milk?

Answer 1

To find the equilibrium temperature of the banana and milk, we can use the principle of conservation of energy. The heat lost by the banana is equal to the heat gained by the milk.

We can use the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Let's calculate the heat lost by the banana:

Q_banana = m_banana * c_banana * ΔT_banana

where:
m_banana = 0.250 kg (mass of the banana)
c_banana = 3520 J/kgK (specific heat of the banana)
ΔT_banana = (equilibrium temperature - 17°C) (change in temperature of the banana)

Now, let's calculate the heat gained by the milk:

Q_milk = m_milk * c_milk * ΔT_milk

where:
m_milk = 0.350 kg (mass of the milk)
c_milk = 4200 J/kgK (specific heat of the milk)
ΔT_milk = (equilibrium temperature - 8°C) (change in temperature of the milk)

Since the heat lost by the banana is equal to the heat gained by the milk, we can set up the equation:

Q_banana = Q_milk

m_banana * c_banana * ΔT_banana = m_milk * c_milk * ΔT_milk

Substituting the given values:

0.250 kg * 3520 J/kgK * (equilibrium temperature - 17°C) = 0.350 kg * 4200 J/kgK * (equilibrium temperature - 8°C)

Now we can solve for the equilibrium temperature:

0.250 kg * 3520 J/kgK * equilibrium temperature - 0.250 kg * 3520 J/kgK * 17°C = 0.350 kg * 4200 J/kgK * equilibrium temperature - 0.350 kg * 4200 J/kgK * 8°C

0.250 kg * 3520 J/kgK * equilibrium temperature - 0.250 kg * 3520 J/kgK * 17°C = 0.350 kg * 4200 J/kgK * equilibrium temperature - 0.350 kg * 4200 J/kgK * 8°C

Now, we can simplify and solve for the equilibrium temperature.

(0.250 kg * 3520 J/kgK - 0.350 kg * 4200 J/kgK) * equilibrium temperature = 0.250 kg * 3520 J/kgK * 17°C - 0.350 kg * 4200 J/kgK * 8°C

(0.250 kg * 3520 J/kgK - 0.350 kg * 4200 J/kgK) * equilibrium temperature = 0.250 kg * 3520 J/kgK * 17°C - 0.350 kg * 4200 J/kgK * 8°C

(0.220 kg * 3520 J/kgK) * equilibrium temperature = (0.250 kg * 3520 J/kgK * 17°C - 0.350 kg * 4200 J/kgK * 8°C)

Now, let's calculate the right side of the equation:

(0.220 kg * 3520 J/kgK) * equilibrium temperature = (0.250 kg * 3520 J/kgK * 17°C - 0.350 kg * 4200 J/kgK * 8°C)

0.220 kg * 3520 J/kgK * equilibrium temperature = (0.250 kg * 3520 J/kgK * 17°C - 0.350 kg * 4200 J/kgK * 8°C)

0.220 kg * 3520 J/kgK * equilibrium temperature = (4400 J - 1176 J)

0.220 kg * 3520 J/kgK * equilibrium temperature = 3224 J

Finally, we can solve for the equilibrium temperature:

equilibrium temperature = 3224 J / (0.220 kg * 3520 J/kgK)

equilibrium temperature = 3224 J / 774.4 J/K

equilibrium temperature = 4.16°C

Therefore, the equilibrium temperature of the banana and milk is approximately 4.16°C.