Question

A copper calorimeter can with mass 0.100kg contains 0.160kg of water and 0.018kg of ice in thermal equilibrium at atmospheric pressure.

Part A
If 0.750kg of lead at a temperature of 255 c is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings

Answer 1

Step-by-step explanation:

The given data is as follows.

Mass of copper calorimeter, (
`m_(cu)`) = 0.1 kg

specific heat of copper calorimeter, (
`c_(cu)`) = 390 J/kg K

mass of water, (
`m_(w)`) = 0.160 kg

specific heat of water, (
`c_(w)`) = 4190 J/kg K

mass of ice, (
`m_i`) = 0.018 kg

latent heat of ice =
`334 * 103 J/kg`

mass of lead, (
`m_(Pb)`) = 0.75 kg

specific heat of lead, (
`c_(Pb)`) = 130 J/kg K

Heat lost by the lead is

Q =
`m_(pb) * C_(Pb) * (255 - T)`

=
`0.75 * 130 J/kg K * (255 - T)`

= 97.5 (255 - T)

Now, heat gained by calorimeter is calculated as follows.

Q =
`(m_(w) + m_(ice)) * c_(w) * T + m_(Cu) * C_(Cu) * T + m_(ice) * L_(f)`

=
`(0.160 + 0.018)(4190) T + (0.1)(390 J/kg K) T + (0.018 kg )(334 * 103)`

Hence, heat loss by lead is equal to heat gained by calorimeter .

So, 97.5 (255 - T ) = 745.82 T + 39 T + 6012

18850.5 = 882.32 T

T =
`21.36^(o)C`

Thus, we can conclude that the value of final temperature is
`21.36^(o)C`.