An ice cube with a mass of 40.0 g melts in water originally at 25.0 degree C.
a. How much heat does the ice cube absorb from the water when it melts. Report the answer in calories, kilocalories, joules
b. Calculate the number of grams of water that can be cooled to 0 degreeC by the melting ice cube.
a. To find out how much heat the ice cube absorbs from the water when it melts, we can use the formula:
Q = m * ΔH
where,
Q = heat absorbed (in joules)
m = mass of the ice cube (in grams)
ΔH = enthalpy of fusion (or heat of fusion) of water (in calories/gram)
First, let's convert the mass of the ice cube to kilograms:
m = 40.0 g = 0.04 kg
The enthalpy of fusion of water is 79.7 calories/gram.
Now we can calculate the heat absorbed:
Q = 0.04 kg * 79.7 cal/g = 3.188 cal
To convert calories to kilocalories, we divide by 1000:
Q = 3.188 cal / 1000 = 0.003188 kcal
To convert calories to joules, we multiply by 4.184:
Q = 3.188 cal * 4.184 J/cal = 13.35 J
So, the ice cube absorbs approximately 3.188 calories, 0.003188 kilocalories, and 13.35 joules of heat from the water when it melts.
b. To calculate the number of grams of water that can be cooled to 0 degrees Celsius by the melting ice cube, we can use the formula:
Q = m * c * ΔT
where,
Q = heat exchanged (in joules)
m = mass of the water (in grams)
c = specific heat capacity of water (1 cal/g°C or 4.184 J/g°C)
ΔT = change in temperature (25.0°C - 0°C = 25.0°C)
We know that the heat absorbed by the ice cube (Q) is equal to the heat lost by the water, so:
Q = 13.35 J = m * c * ΔT
Rearranging the equation to solve for m:
m = Q / (c * ΔT)
m = 13.35 J / (4.184 J/g°C * 25.0°C)
m ≈ 0.127 g
Therefore, approximately 0.127 grams of water can be cooled to 0 degrees Celsius by the melting ice cube.
a. To find the amount of heat absorbed by the ice cube when it melts, we can use the formula:
Q = mass * specific heat * change in temperature
First, let's find the mass of the ice cube in grams:
Given: Mass of ice cube = 40.0 g
Now, let's find the change in temperature. The ice cube will melt at 0 degree Celsius, so the change in temperature is:
Change in temperature = 0 - 25 = -25 degrees Celsius
Next, let's find the specific heat of ice. The specific heat of ice is 0.50 cal/g°C, which means it takes 0.50 calories to raise the temperature of 1 gram of ice by 1 degree Celsius.
Now we can plug the values into the formula:
Q = 40.0 g * 0.50 cal/g°C * (-25°C)
Calculating this expression, we get:
Q = -500.0 cal
Since heat is being absorbed by the ice cube, we consider its value as positive:
Q = 500.0 cal
To convert calories to kilocalories, we divide by 1000:
500.0 cal / 1000 = 0.5 kcal
To convert calories to joules, we use the conversion factor 1 cal = 4.184 J:
Q = 0.5 kcal * 4.184 J/cal
Calculating this expression, we get:
Q = 2.092 J
So, the amount of heat absorbed by the ice cube when it melts is 500.0 calories, 0.5 kilocalories, or 2.092 joules.
b. To calculate the number of grams of water that can be cooled to 0 degrees Celsius by the melting ice cube, we can use the formula:
Q = mass * specific heat * change in temperature
Since we want to find the mass, we rearrange the formula as:
mass = Q / (specific heat * change in temperature)
Given the following values:
Q (heat absorbed by ice cube) = 500.0 cal
specific heat of water = 1.00 cal/g°C
change in temperature = 0 - 25 = -25 degrees Celsius
Plugging these values into the formula:
mass = 500.0 cal / (1.00 cal/g°C * (-25°C))
Calculating this expression, we get:
mass = 20.0 g
Thus, the melted ice cube can cool 20.0 grams of water to 0 degrees Celsius.
q = mass ice x heat fusion
mass ice x heat fusion + [mass water x specific heat water x (Tfinal-Tinitial)] = 0