Respuesta :
Answer:
[tex]m_a=931.9955\ g[/tex]
Explanation:
Given:
- initial temperature of Aluminium, [tex]T_a_i=91.4^{\circ}C[/tex]
- mass of water, [tex]m=200\ g[/tex]
- initial temperature of water, [tex]T_w_i=15.5^{\circ}C[/tex]
- specific heat of aluminium, [tex]c_a=0.897\ J.g^{-1}.^{\circ}C^{-1}[/tex]
- specific heat of water, [tex]c_w=4.18\ J.g^{-1}.^{\circ}C^{-1}[/tex]
- final temperature of the mixture, [tex]T_f=18.9^{\circ}C[/tex]
Heat lost by Al to go from 91.4°C to 18.9°C:
[tex]Q_a=m_a.c_a.\Delta T_a[/tex]
where:
[tex]\Delta T_a=[/tex] temperature difference for Al
[tex]m_a=[/tex] mass of aluminium
[tex]\Rightarrow Q_a=m_a\times 0.897\times 72.5[/tex]
[tex]Q_a=65.0325\times m_a[/tex] ...........................................(1)
Heat gained by water to go from 15.5°C to 18.9°C:
[tex]Q_w=m_w.c_w.\Delta T_w[/tex]
where:
[tex]\Delta T_w=[/tex] temperature difference for water
[tex]\Rightarrow Q_w=200\times 4.18\times 72.5[/tex]
[tex]Q_w=60610\ J[/tex] ...........................................(2)
According to the law of conservation of energy:
When there is no heat loss to the surrounding:
[tex]Q_a=Q_w[/tex]
[tex]65.0325\times m_a=60610[/tex]
[tex]m_a=931.9955\ g[/tex]
