Question

52.76 g of copper pellets are removed from a 333^°C oven and immediately dropped into 151 mL of water at 16^°C in an insulated cup. What will the new water temperature be?
Specific heat of the copper is 385 J/kg·^°C, specific heat of the water is 4190 J/kg·^°C.

Answer 1

the heat lost by the copper pellets will be transferred to the water, since it is an insulated cup.

-Q_copper=Q_water
MC*(-dT)=mc(dt)
We use 385 J/(kg*K) for specific heat of copper and 4190J/(kg*K) for specific heat of water.
151mL will need to converted to m^3 then to mass.
151mL=0.000151m^3
density=mass/volume, density of water = 1000kg/m^3
1000=m/0.000151
m=0.151 kg.
52.76 g also needs to be converted to 0.0528 kg
We are ready to use equation.
-(0.0528)*(385)*(T-333)=
(0.151)*(4190)*(T-16)
T=25.86 degree Celsius.