A thermos contains 150cm3 of coffee at 85° C. To cool the coffee, you drop two 10g ice cubes into the thermos. The ice cubes are initially at 0° C and melt completely. What is the final temperature of the coffee? Treat the coffee as if it were water. (See Table 12.2 and Table 12.3 for appropriate constants.)

Please HELP!!!!

The heat gained by the ice cubes in melting and rising ti the final temperature equals the energy lost by the coffee.

Write that as an equation and solve for the final temperature, T.

150*(85 - T) = 20*[80 + 1.0*(T - 0)]

12,750 - 150 T = 1600 + 20T
170 T = 11,150
T = ___ deg C

The 80 is the heat of fusion, in cal/g, and the 1.0 is the specific heat of water, in cal/(g*C)

Sure, I'm here to help! But before we start, I have to ask: Why did the coffee file a police report?

Because it got mugged! ☕️👮‍♂️

Now, let's get back to your question. To find the final temperature of the coffee after the ice cubes melt, we can use the principle of conservation of energy. The heat gained by the ice cubes from the coffee will be equal to the heat lost by the coffee.

First, let's calculate the heat gained by the ice cubes. To do that, we need to find the heat required to convert the ice into water. This can be calculated using the heat of fusion equation:

Q = m * ΔHf

where Q is the heat gained, m is the mass of the ice, and ΔHf is the heat of fusion of water. From Table 12.2, the heat of fusion of water is 334 J/g.

We have two ice cubes, each weighing 10g, so the total mass is 20g. Plugging in the values, we have:

Q = 20g * 334 J/g

Calculating that, Q = 6680 J.

Now, let's find the heat lost by the coffee. We can use the specific heat capacity equation:

Q = m * c * ΔT

where Q is the heat lost, m is the mass of the coffee, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.

The mass of the coffee is given as 150cm3, which is equivalent to 150g since the density of water is 1g/cm3.

Plugging in the values, we have:

Q = 150g * 4.18 J/g°C * (85°C - Tf)

Now, we can equate the heat gained by the ice cubes with the heat lost by the coffee:

6680 J = 150g * 4.18 J/g°C * (85°C - Tf)

Solving for Tf, we find:

Tf = 85°C - (6680 J / (150g * 4.18 J/g°C))

Calculating that, we get Tf ≈ 78.8°C.

So the final temperature of the coffee after the ice cubes melt is approximately 78.8°C. Your coffee is ready to be enjoyed, but don't forget to bring a spoonful of laughter to every sip! 😄

To find the final temperature of the coffee, we need to calculate the heat lost by the coffee and the heat gained by the ice cubes during the cooling process.

First, let's calculate the heat lost by the coffee using the formula:

Q1 = m1 * c * ΔT1

Where:
Q1 = heat lost by the coffee
m1 = mass of the coffee (in grams)
c = specific heat capacity of water (in J/g °C)
ΔT1 = change in temperature of the coffee (initial temperature - final temperature)

Given:
m1 = 150 cm^3 = 150 g (since 1 cm^3 of water is equivalent to 1 gram)
c = 4.18 J/g °C (specific heat capacity of water)
initial temperature of the coffee (T1) = 85 °C (given)
final temperature of the coffee (T2) = ?
ΔT1 = T1 - T2

Now, let's calculate the heat gained by the ice cubes using the formula:

Q2 = m2 * Lf

Where:
Q2 = heat gained by the ice cubes
m2 = mass of the ice cubes (in grams)
Lf = latent heat of fusion of water (in J/g)

Given:
m2 = 2 * 10 g (since there are two 10g ice cubes)
Lf = 334 J/g (latent heat of fusion of water)

Since the heat lost by the coffee is equal to the heat gained by the ice cubes, we can set Q1 = Q2.

m1 * c * ΔT1 = m2 * Lf

Plugging in the given values, we have:

150 * 4.18 * (85 - T2) = 2 * 10 * 334

Now, let's solve for T2:

627.6 * (85 - T2) = 6680

627.6 * 85 - 627.6 * T2 = 6680

53346 - 627.6 * T2 = 6680

-627.6 * T2 = 6680 - 53346

-627.6 * T2 = -46666

T2 = (-46666) / (-627.6)

T2 ≈ 74.33 °C

Therefore, the final temperature of the coffee after dropping the ice cubes is approximately 74.33 °C.

To find the final temperature of the coffee, we can use the principle of conservation of energy. The heat lost by the coffee as it cools down is equal to the heat gained by the ice cubes as they melt. We can express this relationship using the equation:

(heat lost by coffee) = (heat gained by ice cubes)

The heat lost by the coffee is calculated using the equation:

q = m * c * ΔT

Where:
q is the heat lost by the coffee,
m is the mass of the coffee,
c is the specific heat capacity of water,
ΔT is the change in temperature.

The heat gained by the ice cubes is calculated using the equation:

q = m * ΔH_f

Where:
q is the heat gained by the ice cubes,
m is the mass of the ice cubes,
ΔH_f is the heat of fusion (latent heat) of water, which is the amount of heat required to convert one gram of a substance from the solid to the liquid phase.

Now, let's calculate the heat lost by the coffee:

m_coffee = 150 cm^3 = 150 g (since the density of water is 1 g/cm^3)
c_water = 4.18 J/g·°C (specific heat capacity of water)
ΔT = final temperature - initial temperature = T_f - 85°C

q_coffee = m_coffee * c_water * ΔT

Now, let's calculate the heat gained by the ice cubes:

m_ice = 2 * 10 g = 20 g (since there are two ice cubes)
ΔH_f = 334 J/g (heat of fusion of water)

q_ice = m_ice * ΔH_f

Since the heat lost by the coffee is equal to the heat gained by the ice cubes, we can set up the equation:

q_coffee = q_ice

m_coffee * c_water * ΔT = m_ice * ΔH_f

Now, we can solve this equation for the final temperature, T_f:

T_f = (m_ice * ΔH_f) / (m_coffee * c_water) + 85°C

Substituting the known values:

T_f = (20 g * 334 J/g) / (150 g * 4.18 J/g·°C) + 85°C

Calculating this expression will give the final temperature of the coffee after adding the ice cubes.