A 40-g block of ice is cooled to
−71°C
and is then added to 590 g of water in an 80-g copper calorimeter at a temperature of 27°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 0°C, melt, and then continue warming as water. (The specific heat of ice is 0.500 cal/g · °C = 2,090 J/kg · °C.)
Tf = ? (°C)
mass of ice final (g)
For the ice to reach 0 degrees, you will need Q = c• m• ΔT 0.04• 2090 • 71 = 5935,6 J
The latent heat needed to melt the ice is 334 000 J / kg.
So to melt the ice, you will need 334 000 • 0.040 = 13360 J
Assume final temperature = x degC
Specific heat of water is 4180 J / k g degC.
So heat applied to ice will be 0.04• 4180• x = 1672 x
Total heat energy applied = 5935,6 + 13360 + 167,2x
Heat from water =0.590 • 4180 • (27-x) = 2466.2• (27-x)
Specific heat of copper is 385 J / kg degC.
Heat from copper = 0.08• 385 • (27-x) = 30.8• (27-x)
Heat gained = heat loss
5935,6 + 13360+167.2 x =2466.2(27-x) +30.8• (27-x)
19296 +167.2 x = 2466.2• (27-x) +30.8• (27-x)
x =8.53
So final temperature is 8.53 degC.
To determine the final temperature of the system, we need to consider the heat gained or lost by each component.
1. Heat lost by the ice:
The heat lost by the ice is given by the formula: Q = m * c * ΔT
where:
- Q is the heat lost (in joules or calories)
- m is the mass of the ice (in grams)
- c is the specific heat of the ice (in calories/gram°C or joules/gram°C)
- ΔT is the change in temperature (in °C)
Using this formula, we can calculate the heat lost by the ice as it warms from -71°C to 0°C:
Q_ice = (40 g) * (0.500 cal/g °C) * (0°C - (-71°C))
2. Heat gained by the water:
The heat gained by the water is given by the formula: Q = m * c * ΔT
where:
- Q is the heat gained (in joules or calories)
- m is the mass of the water (in grams)
- c is the specific heat of water (in calories/gram°C or joules/gram°C)
- ΔT is the change in temperature (in °C)
Using this formula, we can calculate the heat gained by the water as it warms from 27°C to the final temperature:
Q_water = (590 g) * (1.00 cal/g °C) * (Tf - 27°C)
3. Heat gained by the copper calorimeter:
The heat gained by the calorimeter is also given by the formula: Q = m * c * ΔT
where:
- Q is the heat gained (in joules or calories)
- m is the mass of the calorimeter (in grams)
- c is the specific heat of copper (0.092 cal/g °C or 390 J/kg °C)
- ΔT is the change in temperature (in °C)
Using this formula, we can calculate the heat gained by the calorimeter as it warms from 27°C to the final temperature:
Q_calorimeter = (80 g) * (0.092 cal/g °C) * (Tf - 27°C)
4. Heat required to melt the remaining ice:
If all the ice does not melt and there is some left, we need to calculate the heat required to melt the remaining ice. The formula to calculate this is: Q = m * ΔH_fusion
where:
- Q is the heat required (in joules or calories)
- m is the mass of the remaining ice (in grams)
- ΔH_fusion is the heat of fusion for ice (79.7 cal/g or 334 J/g)
Using this formula, we can calculate the heat required to melt the remaining ice:
Q_melt = (mass of remaining ice g) * (79.7 cal/g)
5. Equating the heat lost and gained:
Since energy is conserved, the total heat lost by the ice must be equal to the total heat gained by the water and calorimeter:
Q_ice = Q_water + Q_calorimeter + Q_melt
6. Solving for the final temperature:
We can now rearrange the equation and solve for the final temperature (Tf).
Note: Make sure to convert all temperatures to Celsius and use the appropriate units (calories or joules) based on the specific heat values given.
I hope this information helps you with your calculations. Let me know if you need any further assistance.
To solve this problem, we need to consider the heat transfer between the substances and apply the principle of energy conservation. We can break down the process into three steps:
Step 1: Heating the ice to its melting point.
Step 2: Melting the ice.
Step 3: Heating the resulting water to the final temperature.
Let's calculate the amount of heat absorbed or released during each step.
Step 1: Heating the ice to its melting point.
To warm the ice from -71°C to 0°C, we need to calculate the heat transfer using the specific heat capacity of ice:
Heat_1 = mass_ice * specific_heat_ice * temperature_change_1
Given:
mass_ice = 40 g
specific_heat_ice = 0.500 cal/g · °C = 2,090 J/kg · °C (convert cal/g · °C to J/g · °C)
temperature_change_1 = 0°C - (-71°C) = 71°C
Plugging in the values:
Heat_1 = 40 g * 2,090 J/g · °C * 71°C
Step 2: Melting the ice.
To melt the ice at 0°C, we need to calculate the heat transfer using the enthalpy of fusion of ice:
Heat_2 = mass_ice * heat_of_fusion
Given:
heat_of_fusion = 334 J/g
Plugging in the values:
Heat_2 = 40 g * 334 J/g
Step 3: Heating the resulting water to the final temperature.
To heat the resulting water from the melting point to the final temperature, we need to calculate the heat transfer using the specific heat capacity of water:
Heat_3 = mass_water * specific_heat_water * temperature_change_2
Given:
mass_water = 590 g
specific_heat_water = 1 cal/g · °C = 4,184 J/kg · °C (convert cal/g · °C to J/g · °C)
temperature_change_2 = Tf - 0°C (as water)
Plugging in the values:
Heat_3 = 590 g * 4,184 J/g · °C * (Tf - 0°C)
Now, according to the principle of energy conservation, the total heat absorbed in the system must equal zero:
Heat_1 + Heat_2 + Heat_3 = 0
Substituting the calculated values, we can solve for Tf:
40 g * 2,090 J/g · °C * 71°C + 40 g * 334 J/g + 590 g * 4,184 J/g · °C * (Tf - 0°C) = 0
Simplifying the equation will give us the final temperature (Tf) of the system.