Answer:[tex]T=69.56^{\circ}[/tex]
Explanation:
Given
volume of [tex]V=144 cm^3[/tex]
mass of coffee[tex]=\rho \times volume =1\times 144=144 gm[/tex]
Initial temperature of coffee[tex]=95^{\circ}[/tex]
mass of ice cubes =10.9 gm
and latent heat of ice=0.336 MJ
heat absorb by ice lowers coffee temperature
[tex]m_{ice}\times L=m_{coffee}\times c\times (95-T)[/tex]
[tex]10.9\times 0.336\times 10^6=144\times 4.184\times (95-T)[/tex]
[tex]336=13.21\times (95-T)[/tex]
[tex]T=69.56^{\circ}[/tex]
