Answer:
[tex]C_m=0.474\frac{J}{g\°C}[/tex]
Explanation:
Hello.
In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:
[tex]Q_m=-Q_w[/tex]
Thus, in terms of masses, specific heats and temperatures we can write:
[tex]m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)[/tex]
Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:
[tex]C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}[/tex]
Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:
[tex]C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}[/tex]
Best regards!
The specific heat capacity of the metal is 0.474 J/gºC
We'll begin by calculating the mass of the water.
Density = 1 g/mL
Volume = 80 mL
Mass = Density × Volume
Mass of water = 1 × 80
Mass of water (M) = 80 g
Initial temperature of water (T₁) = 24 °C
Final temperature (T₂) = 28.4 °C
Change in temperature (ΔT) = T₂ – T₁ = 28.4 – 24 = 4.4 °C
Specific heat capacity of water (C) = 4.184 J/gºC
Q = MCΔT
Q = 80 × 4.184 × 4.4
Heat gained = 1472.768 J
Heat lost = –1472.768 J
Mass of water (M) = 44 g
Initial temperature of water (T₁) = 99 °C
Final temperature (T₂) = 28.4 °C
Change in temperature (ΔT) = T₂ – T₁ = 28.4 – 99 = –70.6 °C
Q = MCΔT
–1472.768 = 44 × C × –70.6
–1472.768 = –3106.4 × C
Divide both side by –3106.4
C = –1472.768 / –3106.4
Therefore, the specific heat capacity of the metal is 0.47 J/gºC
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