Question

A 20.64 g ice cube at -1.93 ^oC is place in an aluminum cup whose initial temperature is 137.9 ^oC. The system comes to an equilibrium temperature of 17.96 ^oC. What is the mass in grams of the cup? The specific heat of ice is 2.093 J/g ^oC and that of aluminum is 0.901 J/g ^oC.

Answer 1

heat gained by ice Q1 = M_ice*dT1 + Mice*Lf +Mice*Cw*dT2

Q1 = (20.64*2.093*1.93)+(20.64*334)+(20.64*4.19*17.96)

Q1 = 8530.345 J

heat loss by cup = M_cup*Cal*dT

Q2 = Mcup*0.901*(1379-17.96) = 1226.29704*M_cup

from calorimetry

Q2 = Q1

1226.29704*M_cup = 8530.345

Mcup = 6.956 g

Answer 2

use the equation q=(m*C*deltaT),

where m equals the mass,

C equals the specific heat,

deltaT equals the change in temperature, and

q equals the heat