Q
a piece of titanium metal with a mass of 20.8
PaulBirrell Release Time: 3月 22, 2024, 6:12 下午
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g is heated to 98.6°C and then dropped into a beaker containing 35.0 g of water at 25.0°C. The heat capacity of titanium is 0.523 J/g°C and the heat capacity of water is 4.18 J/g°C. Firstly, note that when the titanium is dropped into the water, it will lose heat to the water, causing the water to heat up. The final temperature will be somewhere between 98.6°C and 25.0°C. The law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. Thus, when the hot piece of titanium is dropped into the cooler water, the heat (energy) lost by the titanium should equal the heat gained by the water. We can write this as: heat lost by titanium = heat gained by water Which simplifies to: mass_titanium * C_titanium * ΔT_titanium = mass_water * C_water * ΔT_water Where ΔT is the change in temperature (in this case, the final temperature minus the initial temperature), and C represents the heat capacity of each substance. We're trying to solve for the final temperature (T_f), so let's modify the equations like so: 20.8 g * 0.523 J/g°C * (98.6°C - T_f) = 35.0 g * 4.18 J/g°C * (T_f - 25.0°C) Now you can solve for T_f indirectly. But let's make it straight forward, Let's call the left hand side of the equation LHS and the right hand side RHS. LHS = 20.8 g * 0.523 J/g°C * (98.6°C) - 20.8 g * 0.523 J/g°C * T_f .......(Eqn 1) RHS = 35.0 g * 4.18 J/g°C * T_f - 35.0 g * 4.18 J/g°C * 25.0°C ........(Eqn 2) Now, T_f = (LHS + 35.0 g * 4.18 J/g°C * 25.0°C ) / (20.8 g * 0.523 J/g°C + 35.0 g * 4.18 J/g°C) So the only thing you need to do is to calculate (LHS + 35.0 g * 4.18 J/g°C * 25.0°C ) and (20.8 g * 0.523 J/g°C + 35.0 g * 4.18 J/g°C) and divide them. There is a lot of calculation to be done but eventually you will get the value of T_f. Please bear with any typing error in calculations. So, be sure to check your math.
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