5) An insulated Thermos contains 190 cm3 of hot coffee at 79.0°C. You put in a 11.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg•K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3.
but i cant do it!!!
Change all the temp. to Kelvin first. Ice is 0C in solid form so there is a latent heat too.
4187(190/1000)(352-T)=333x1000x11/1000 + 4187(11/1000)(T-273)
Solve for T?
Then (T-273) is in Celsius.
Difference = 79 - (T-273)
The answer is 79.909 degrees C, but don't know how to get that
To solve this problem, we need to calculate the heat gained by the ice cube and the heat lost by the coffee. First, let's calculate the heat gained by the ice cube using the formula:
Q = m * L
Where:
Q is the heat gained by the ice cube,
m is the mass of the ice cube, and
L is the latent heat of fusion.
Given:
m = 11.0 g (mass of the ice cube)
L = 333 kJ/kg (latent heat of fusion)
We need to convert the mass of the ice cube from grams to kilograms:
m = 11.0 g ÷ 1000 = 0.011 kg
Now, let's calculate the heat gained by the ice cube:
Q = 0.011 kg * 333 kJ/kg = 3.663 kJ
Next, let's calculate the heat lost by the coffee using the formula:
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 of water, and
ΔT is the change in temperature of the coffee.
Given:
c = 4186 J/kg∙K (specific heat of water)
m = 190 cm3 (volume of the coffee)
Density of water = 1.00 g/cm3
The volume of the coffee is given in cubic centimeters, so we need to convert it to kilograms:
m = 190 cm3 * 1.00 g/cm3 ÷ 1000 = 0.190 kg
Since the coffee temperature decreases, ΔT is negative.
Let's calculate ΔT using this formula:
ΔT = Q / (m * c)
ΔT = -3.663 kJ / (0.190 kg * 4186 J/kg∙K)
Now, let's substitute the values and solve for ΔT:
ΔT = -3.663 kJ / (0.190 kg * 4186 J/kg∙K) = -8.39 K
Note that the change in temperature is negative since the coffee cools down.
Therefore, the coffee has cooled by approximately 8.39 degrees Celsius once the ice has melted and equilibrium is reached.
Are you supposed to assume that the ice is intially 0 C? It could be lower.
The heat gained by the 11 g of ice, as it melts and rises to final temperature T, is equal to the heat lost by the 190 g of original liquid water, as it cools from 79 C to T.
Write that as an equation and solve for T. .