In: Physics
25.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 18.0 ∘C in an insulated cup.
1.)What will the new water temperature be?(in Celsius
Equation for heat:
q = m · Cp · ΔT
where,
q = heat
m = mass
Cp = specific heat
ΔT = (T higher – T lower)
The trick for these type of problems, which
involves substances being brought into contact,
is to make all ΔT positive; ΔT = T higher - T lower.
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q lost from hot pellets = q gained in cool water
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q lost from hot pellets:
q pellets = m · Cp · ΔT
where,
m = mass Cu = 25.0 g
Cp Cu = 385 J/kg·K = 385 J/kg·ºC x (1kg / 1,000g) = 0.385
J/kg·ºC
Ti = 300.ºC
ΔT = (T higher – T lower) = (Ti – Tf) = (300.ºC – Tf)
Substituting numbers into the equation:
q pellets = (25.0 g) · (0.385 J/g·ºC) · (300.ºC – Tf)
==> q pellets = 2887.5 J – 9.625 J/ºC · Tf ~(eq1)
q gained in cool water:
q water = m · Cp · ΔT
where,
assuming density of water = 1g/mL
m water = 120.mL x (1g/mL) = 120.g
Cp water = 4190 J/kg·K = 4190 J/kg·ºC x (1kg / 1,000g) = 4.190
J/g·ºC
Ti = 18.0ºC
ΔT = (T higher – T lower) = (Tf – Ti) = (Tf – 18.0ºC)
Substituting numbers into the equation:
q water = (120.g) · (4.190 J/g·ºC) · (Tf – 18.0ºC)
==> q water = 502.8 J/ºC · Tf – 9050.4 ~(eq2)
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q lost from hot pellets = q gained in cool water
==> ~(eq1) = ~(eq2)
2887.5 J – 9.625 J/ºC · Tf = 502.8 J/ºC · Tf – 9050.4 J
or, 512.425 J/ºC · Tf = 11,937.9 J
or, Tf =23.29 ºC
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