a 1.00kg piece of aluminum metal at 90 degrees Celsiusis placed in 4 L (=4.00kg)of water at 25°C determine the final temperature

By applying the Law of Conservation of Energy, the final temperature of water is equal to 28.63°CGiven the following data:Mass of aluminum metal = 1.00 kgVolume of water = 4 LMass of water = 4.00 kgTemperature of aluminum metal = 90.0°CInitial temperature of water = 25.0°CScientific data:Specific heat capacity of water = 4200 J/kg°CSpecific heat capacity of aluminum = 900 J/kg°CMathematically, quantity of heat energy is given by the formula;[tex]Q=mc\theta[/tex]Where:Q represents the quantity of heat energy.m represents the mass of an object.c represents the specific heat capacity.∅ represents the change in temperature.According to the Law of Conservation of Energy, we have: The quantity of heat energy lost by the aluminum metal = The quantity of heat energy gained by the water.[tex]Q_{lost} = Q_{gained}\\\\m_ac_a\theta_a = m_wc_w\theta_w\\\\1(900)(90 - T_f) = 4(4200)(T_f - 25)\\\\81000-900T_f =16800T_f - 420000\\\\16800T_f+900T_f=420000 + 81000\\\\T_f=\frac{501000}{17500} \\\\T_f=28.63[/tex]Final temperature = 28.63°CRead more: https://brainly.com/question/16735503

From the basic "heat lost by hot object=heat gained by colder object" principle, we havem1c1ΔT1=m2c2ΔT2where m1= 1kgm2=4kgc1=900J/kg kc2=4200J/kg k With this information at hand we havem1c1(90-T)=m2c2(T-25)after substituting the given values we can find thatT=28.3^{0}c