Question

30.0g of copper pellets are removed from a 300°C oven and immediately dropped into 110mL of water at 24.0°C in an insulated cup. What will the new temperature be? Please provide full solutions.

Answer 1

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.

--------------------------------------...
q lost from hot pellets = q gained in cool water

--------------------------------------...
q lost from hot pellets:

q pellets = m · Cp · ΔT
where,
m = mass Cu = 30.0 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 = (30.0 g) · (0.385 J/g·ºC) · (300.ºC – Tf)

==> q pellets = 3,465 J – 11.50 J/ºC · Tf ~(eq1)

--------------------------------------...
q gained in cool water:

q water = m · Cp · ΔT
where,
assuming density of water = 1g/mL
m water = 110.mL x (1g/mL) = 110.g
Cp water = 4190 J/kg·K = 4190 J/kg·ºC x (1kg / 1,000g) = 4.190 J/g·ºC
Ti = 24.0ºC
ΔT = (T higher – T lower) = (Tf – Ti) = (Tf – 24.0ºC)

Substituting numbers into the equation:
q water = (110.g) · (4.190 J/g·ºC) · (Tf – 24.0ºC)

==> q water = 461 J/ºC · Tf – 10056~(eq2)

--------------------------------------...
q lost from hot pellets = q gained in cool water

==> ~(eq1) = ~(eq2)

3,465 J – 11.50 J/ºC · Tf = 461 J/ºC · Tf – 10056 J

Solving for Tf:

472.5 J/ºC · Tf = 13521 J

Tf = 13521J / 472.5 J/ºC
= 28.615°C