Answer:
Approximately 73 grams if no heat is lost in this process.
Explanation:
Relevant data and their sources:
The heat it takes to raise the calorimeter and its content from 0°C to 40°C includes three parts:
Note that to avoid rounding errors, most calculations in this explanation do not follow significant figure rules.
The latent heat of fusion of water gives the amount of heat required to melt each unit mass of ice at 0°C to water without changing the temperature. (As a side note, that's also the amount heat released when unit mass of water turns into ice at 0°C.)[tex]L_\mathrm{F}(\text{Water}) = \rm 334\;J\cdot g^{-1}[/tex]. Melting each gram of ice requires approximately 334 joules of heat, so melting 10 grams of ice will require about 3,340 joules of heat.
There's no state change involved after all that 10 grams of ice melted into water. There's now 80 grams of water at 0°C to melt. The specific heat of water gives the amount of energy required to raise one gram of water by one degree Celsius (which is the same as one degree Kelvin.) [tex]c(\text{Water}) = \rm 4.179\;J\cdot g^{-1}[/tex].
[tex]Q(\text{water}) = c(\text{water})\cdot m(\text{water}) \cdot \Delta T = \rm 4.179\times 80\times 40 = 13372.8\; J[/tex].
The melting point of copper is approximately 1,100°C, which is much higher than 100 °C. As a result, there will not be any state change in this process. [tex]c(\text{Copper}) = \rm 0.385\;J\cdot g^{-1}[/tex]. Similarly,
[tex]Q(\text{copper}) = c(\text{copper})\cdot m(\text{copper}) \cdot \Delta T = \rm 0.385\times 120\times 40 = 1848\; J[/tex].
Combining all three parts:
[tex]\rm 3340 + 13372.8 + 1848 = 18560.8\; J[/tex].
The steam added to the calorimeter will first condense to form water of 100°C before cooling down to 40°C. The energy released in that process will come in two parts:
To simplify the calculations, let the mass of steam required be [tex]x[/tex] grams. [tex]x>0[/tex].
Similar to the latent heat of fusion of water, the latent heat of vaporization of water gives the amount of heat released for each unit mass of steam that condenses to water at 100°C without change in temperature. (As a side note, that's also the amount of energy required to turn unit mass of water into steam at 100°C.) [tex]L_\mathrm{V}(\text{Water}) = \rm 2.260\;J\cdot g^{-1}[/tex]. In other words, each gram of steam that condenses at 100°C will release approximately 2.260 joule. [tex]x[/tex] grams of steam that condenses at 100°C will release [tex]2.260x[/tex] joules of heat.
Similar to the heat required to heat 80 grams of water from 0°C to 40°C, the heat released when [tex]x[/tex] grams of steam condenses to water can also be found with the specific heat capacity of water. [tex]c(\text{Water}) = \rm 4.179\;J\cdot g^{-1}[/tex].
[tex]Q(\text{water}) = c(\text{water})\cdot m(\text{water}) \cdot \Delta T = \rm 4.178\cdot \mathnormal{x}\times (100 - 40)=250.68\mathnormal{x} \; J[/tex].
[tex]2.260x + 250.68x = \rm 252.94\mathnormal{x}\;J[/tex].
The amount of heat released shall equal to amount of heat absorbed. As a result,
[tex]252.94x = 18560.8[/tex].
[tex]\displaystyle x = \rm\frac{18560.8}{252.94}\approx 73\;g[/tex].
In other words, approximately 73 grams of steam at 100°C will be required to heat that calorimeter and its contents to 40°C.
Answer:
7.4 g
Explanation:
Heat absorbed by copper:
= mc Cc ΔTc
= (120 g) (0.39 J/g/°C) (40°C − 0°C)
= 1872 J
Heat absorbed by ice:
= mi Li + mi Cw ΔTi
= (10 g) (334 J/g) + (10 g) (4.18 J/g/°C) (40°C − 0°C)
= 5012 J
Heat absorbed by water:
= mw Cw ΔTw
= (70) (4.18 J/g/°C) (40°C − 0°C)
= 11704 J
Heat lost by steam:
= ms Ls + ms Cw ΔTs
= m (2264.7 J/g) + m (4.18 J/g/°C) (100°C − 40°C)
= 2515.5 J/g m
1872 J + 5012 J + 11704 J = 2515.5 J/g m
m = 7.39 g
Rounded to two significant figures, the mass of steam needed is 7.4 grams.
