What mass of steam initially at 120degree celcius is needed to warm 200g of water in a 100g glass container from 20.0 degree celcius to 50.0degree celcius?
Answers
Answer:
The steam loses energy in three stages. In the first stage, the steam is cooled to 100^{\circ} \mathrm{C}100∘C. The energy transfer in the process is
Q_{1} =m_{s} c_{s} \Delta T=m_{s}\left(2.01 \times 10^{3} \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(-30.0^{\circ} \mathrm{C}\right)Q1=mscsΔT=ms(2.01×103 J/kg⋅∘C)(−30.0∘C) =-m_{s}\left(6.09 \times 10^{4} \mathrm{~J} / \mathrm{kg}\right)=−ms(6.09×104 J/kg)
Where m_{s}ms is the unknown mass of the steam. In the second stage, the steam is converted to water. To find the energy transfer during this phase change, we use Q=-m L_{0},Q=−mL0, where the negative sign indicates that energy is leaving the steam:
Q_{2}=-m_{s}\left(2.26 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right)Q2=−ms(2.26×106 J/kg)
In the third stage, the temperature of the water created from the steam is reduced to 50.0^{\circ} \mathrm{C}50.0∘C. This change requires an energy transfer of
Q_{3} =m_{s} c_{\omega} \Delta T=m_{s}\left(4.19 \times 10^{3} \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(-50.0^{\circ} \mathrm{C}\right)Q3=mscωΔT=ms(4.19×103 J/kg⋅∘C)(−50.0∘C) =-m_{s}\left(2.09 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right)=−ms(2.09×105 J/kg)
Adding the energy transfers in these three stages, we obtain
Q_{\text {hot. }}= Q_{1}+Q_{2}+Q_{3}Qhot. =Q1+Q2+Q3 =-m_{s}\left(6.03 \times 10^{4} \mathrm{~J} / \mathrm{kg}+2.26 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right).=−ms(6.03×104 J/kg+2.26×106 J/kg). \left.+2.09 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right)+2.09×105 J/kg) =-m_{s}\left(2.53 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right)=−ms(2.53×106 J/kg)
Now, we turn our attention to the temperature increase of the water and the glass. Using Equation 20.4, we find that
Q_{\text {cold }}=(0.200 \mathrm{~kg})\left(4.19 \times 10^{3} \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(30.0^{\circ} \mathrm{C}\right)Qcold =(0.200 kg)(4.19×103 J/kg⋅∘C)(30.0∘C) +(0.100 \mathrm{~kg})\left(837 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(90.0^{\circ} \mathrm{C}\right)+(0.100 kg)(837 J/kg⋅∘C)(90.0∘C) = 2.77 \times 10^{4} \mathrm{~J}=2.77×104 J
Using Equation 20.5, we can solve for the unknown mass:
Q_{\text {cold }} =-Q_{\text {hot }}Qcold =−Qhot 2.77 \times 10^{4} \mathrm{~J} =-\left[-m_{s}\left(2.59 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right)\right]2.77×104 J=−