An aluminum calorimetry cup [mc = 0.6 kg, cc = 0.22 kcal/(kgC)] is filled with water [mw = 2.4 kg, cw = 1 kcal/(kgC)]. It is at a temperature of 30C. A sample of an unknown substance with mass mm = 2.34 kg, originally at a temperature of 210C, is added. After a while, the temperature of the water is 35C. What is the specific heat of this material?
To find the specific heat of the unknown material, we can use the principle of conservation of energy. The heat gained by the water and the cup is equal to the heat lost by the unknown substance.
The equation we can use is:
(mw * cw * ΔTw) + (mc * cc * ΔTc) = mm * cm * ΔTm
where:
mw = mass of water
cw = specific heat of water
ΔTw = change in temperature of water
mc = mass of calorimetry cup
cc = specific heat of calorimetry cup
ΔTc = change in temperature of calorimetry cup
mm = mass of unknown material
cm = specific heat of unknown material
ΔTm = change in temperature of unknown material
In this case, we know the values for mw, cw, ΔTw, mc, cc, ΔTc, and mm. We need to solve for cm.
The change in temperature for the water can be calculated as:
ΔTw = Tf - Ti = 35C - 30C = 5C
The change in temperature for the calorimetry cup can be calculated as:
ΔTc = ΔTw
Now we can substitute these values into the equation and solve for cm:
(mw * cw * ΔTw) + (mc * cc * ΔTc) = mm * cm * ΔTm
(2.4 kg * 1 kcal/(kgC) * 5C) + (0.6 kg * 0.22 kcal/(kgC) * 5C) = 2.34 kg * cm * ΔTm
(12 kcal) + (0.66 kcal) = 2.34 kg * cm * ΔTm
12.66 kcal = 2.34 kg * cm * ΔTm
Dividing both sides by (2.34 kg * ΔTm):
cm = (12.66 kcal) / (2.34 kg * ΔTm)
Since we don't have the specific value for ΔTm, we cannot find the exact value of cm.