Respuesta :
the Calorimetry relationships you can find the amount of water in the calorimeter is m = 21.3 g
given parameters
- Lead mass M = 200.0 g
- Initial lead temperature T₁ = 176.4ºC
- Specific heat of Lead [tex]c_{e Pb}[/tex] = 0.129 J / g ºC
- Sspecific heat of water [tex]c_{e H_2O}[/tex] = 4.186 J / g ºC
- Initial water temperature T₀ = 21.7ºC
- Equilibrium temperature T_f = 56.4ºC
to find
The body of water
Thermal energy is the energy stored in the body that can be transferred as heat when two or more bodies are in contact. Calorimetry is a technique where the energy is transferred between the body only in the form of heat and in this case the thermal energy of the lead is transferred to the calorimeter that reaches the equilibrium that the thematic energy of the two is equal
Q_{ceded} = Q_{absorbed}
Lead, because it is hotter, gives up energy
Q_{ceded} = M c_{e Pb} (T₁ - T_f)
The calorimeter that is colder absorbs the heat
Q_{absrobed} = m c_{e H_2O} (T_f - T₀)
where M and m are the mass of lead and water, respectively, c are the specific heats, T₁ is the temperature of the hot lead, T₀ the temperature of cold water and T_f the equilibrium temperature
M c_{ePb} (T₁ - T_f) = m c_{eH2O} (T_f - T₀)
m = [tex]\frac{ M\ c_{ePb} \ (T_1 - T_f)}{c_{eH_2O} \ (T_f - T_o)}[/tex]
let's calculate
m = [tex]\frac{200 \ 0.129 (176.4-56.4)}{ 4.186 \ (56.4 -21.7)}[/tex]
m = 3096 / 145.25
m = 21.3 g
Using the Calorimetry relationships you can find the amount of water in the calorimeter is:
m = 21.3 g
learn more about calorimetry here:
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