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)
To determine the final temperature of the system and the mass of ice remaining, we can use the principle of energy conservation. The energy gained by the ice in warming up to its melting point, melting, and then warming up as water will equal the energy lost by the water and the calorimeter.
First, let's calculate the energy gained by the ice:
1. Energy gained to warm up the ice to 0°C:
- Mass of ice (m1) = 40 g
- Specific heat of ice (c1) = 0.500 cal/g · °C = 2,090 J/kg · °C
- Initial temperature of the ice (T1) = -71°C
- Final temperature of the ice when it reaches 0°C (T2) = 0°C
- Energy gained (Q1) = m1 * c1 * (T2 - T1)
2. Energy gained to melt the ice:
- Heat of fusion of ice (L) = 79.7 cal/g = 333,550 J/kg
- Energy gained (Q2) = m1 * L
3. Energy gained to warm up the melted ice (as water) from 0°C to the final temperature:
- Mass of melted ice (m2) = mass of water = 590 g
- Specific heat of water (c2) = 1 cal/g · °C = 4,186 J/kg · °C
- Initial temperature of the melted ice (T3) = 0°C
- Final temperature of the system (Tf) = ?
Now let's calculate the energy lost by the water and calorimeter:
4. Energy lost by the water and calorimeter:
- Mass of water (M) = 590 g
- Specific heat of water (Cw) = 1 cal/g · °C = 4,186 J/kg · °C
- Initial temperature of the water and calorimeter (T4) = 27°C
- Final temperature of the system (Tf) = ?
- Energy lost (Q3) = (M + 80 g) * Cw * (T4 - Tf)
Since energy is conserved, we can equate the gains and losses of energy:
Q1 + Q2 + Q3 = 0
Substituting the respective equations for Q1, Q2, and Q3:
m1 * c1 * (T2 - T1) + m1 * L + (M + 80 g) * Cw * (T4 - Tf) = 0
Now, we can plug in the given values:
m1 = 40 g
c1 = 2,090 J/kg · °C
T1 = -71°C
T2 = 0°C
L = 333,550 J/kg
m2 = 590 g
c2 = 4,186 J/kg · °C
T3 = 0°C
M = 590 g
Cw = 4,186 J/kg · °C
T4 = 27°C
Solving the equation will give us the final temperature of the system (Tf).