A 94.7 g sample of silver (s= 0.237 J/(g x Degree C)), initially at 348.25 Degrees C , is added to an insulated vessel containing 143.6 g of water (s=4.18 J/(g x Degree C)), initially at 13.97 Degrees C. At equilibrium, the finial temperature of the metal-water mixture is 22.63 Degrees C. How much heat was absorbed by the water? The heat capacity of the vessel is 0.244 kJ/Degrees C.

heat absorbed by H2O = [mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal x (Tfinal-Tinitial)

Take note the Ccal is given in kJ/C and specific heat H2O is given in J/C. You must change one of them; the easier one is Ccal.

5.20

A 94.7-g sample of silver (s = 0.237 J/(g · °C)), initially at 348.25°C, is added to an insulated vessel containing 143.6 g of water (s = 4.18 J/(g · °C)), initially at 13.97°C. At equilibrium, the final temperature of the metal–water mixture is 22.63°C. How much heat was absorbed by the water? The heat capacity of the vessel is 0.244 kJ/°C.

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To find the amount of heat absorbed by the water, you need to use the equation:

q = m · C · ΔT

where:
q is the heat absorbed by the substance
m is the mass of the substance
C is the specific heat capacity of the substance
ΔT is the change in temperature

First, let's calculate the heat absorbed by the silver:

q_silver = m_silver · C_silver · ΔT_silver

where:
m_silver = mass of silver = 94.7 g
C_silver = specific heat capacity of silver = 0.237 J/(g x °C)
ΔT_silver = change in temperature of silver = final temperature - initial temperature of silver
= 22.63 °C - 348.25 °C

Next, let's calculate the heat absorbed by the water:

q_water = m_water · C_water · ΔT_water

where:
m_water = mass of water = 143.6 g
C_water = specific heat capacity of water = 4.18 J/(g x °C)
ΔT_water = change in temperature of water = final temperature - initial temperature of water
= 22.63 °C - 13.97 °C

Finally, to find the total heat absorbed by the water, we can substitute the values into the equation:

q_total = q_water + q_silver

Solving this equation will give us the answer.