Respuesta :
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
The initial temperature of the piece of copper is 335.6 °C.
Step 1:
Data obtained from the question
Mass of copper (M꜀) = 248 g
Volume of water = 390 mL
Density of water = 1 g/mL
Initial temperature of water (Tᵥᵥ) = 22.6 °C
Equilibrium temperature (Tₑ) = 39.9 °C
Initial temperature of copper (T꜀) =?
Step 2:
Determination of the mass of water
Volume of water = 390 mL
Density of water = 1 g/mL
Mass of water =?
[tex]Density = \frac{mass}{volume}\\\\1 = \frac{mass}{390}[/tex]
Cross multiply
[tex]Mass = 1 * 390[/tex]
Mass of water = 390 g
Step 3:
Determination the initial temperature of the copper.
Mass of copper (M꜀) = 248 g
Mass of water (Mᵥᵥ) = 390 g
Initial temperature of water (Tᵥᵥ) = 22.6 °C
Equilibrium temperature (Tₑ) = 39.9 °C
Initial temperature of copper (T꜀) =?
NOTE:
1. Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
2. Specific heat capacity of copper (C꜀) = 0.385 J/gºC
Heat lost by copper = heat gained by water
[tex]Q_{c} = Q_{w} \\ \ M_{c} C_{c}(T_{c}-T_{e}) = M_{w} C_{w}(T_{e}-T_{w})\\248* 0.385(T_{c}-39.9) = 390*4.184(39.9-22.6)\\95.48(T_{c}-39.9) = 1631.76*17.3\\95.48(T_{c}-39.9) = 28229.448[/tex]
Divide both side by 95.48
[tex]T_{c} - 39.9 = \frac{28229.448}{95.48}\\T_{c} - 39.9 = 295.658[/tex]
Collect like terms
[tex]T_{c} = 295.658 + 39.9[/tex]
T꜀ = 335.6 °C
Therefore, the initial temperature of the piece of copper is 335.6 °C.
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