Answer:
This is a heat balance question with negligible heat losses to the surroundings.
Sum of heat losses = Sum of the heat gains
We are required to find the final temperature of the calorimeter system T1?
we were given
C = specific heat capacities of Copper c, Water w, Ice i and Lead l
m = masses as mc = 0.1kg, mw = 0.16kg, mi=0.018kg, ml=0.75kg,
T = Temperature of the components, Lead Tl= 225C and at ambient pressure, Temp. To of the water, ice and copper = 25C
Since, at thermal equilibrium after adding the Lead, the lead temperature Tl decreases while copper, water and ice temp decrease, so we have respective heat losses 'q' as,
qc + qw + qi + ql =0
which means that
(mc x Cc x ΔTC) + (mw x Cw x ΔTC) + (mi x Ci x ΔTC) + (ml x Cl x ΔTC) = 0
(0.1x390x (T1 - 25)) + (0.16x4190x (T1 - 25)) + (0.018x4190x (T1 - 0)) + (0.750x130x (T1 - 225)) = 0
882.32T1 - 41558 = 0
T1 = 47.1K
Hence, the final temperature of the calorimeter system is 47.1K
