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

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 = $334 \times 103 J/kg$

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 = $m_{pb} \times C_{Pb} \times (255 - T)$

               = $0.75 \times 130 J/kg K \times (255 - T)$

                = 97.5 (255 - T)

Now, heat gained by calorimeter is calculated as follows.

        Q = $(m_{w} + m_{ice}) \times c_{w} \times T + m_{Cu} \times C_{Cu} \times T + m_{ice} \times L_{f}$

           = $(0.160 + 0.018)(4190) T + (0.1)(390 J/kg K) T + (0.018 kg )(334 \times 103)$

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 $21.36^{o}C$.