What is the total heat capacity of the calorimeter?

Given calorimeter with heat capacity 900J/kgC and mass 10kg. Also given is mass m, of water w/c is in thermal equilibrium w/ calorimeter @ 25C. Im asked what mass of ice @ -30 deg. Celsius shall be added in the system so that only half of the....?

• This is a question in our exam and i got no idea how to answer this: Given calorimeter with heat capacity 900J/kgC and mass 10kg. Also given is mass m, of water w/c is in thermal equilibrium w/ calorimeter @ 25C. Im asked what mass of ice @ -30 deg. Celsius shall be added in the system so that only half of the ice's mass will react to achieve thermal equilibrium. Any idea how to solve this? Final temperature i guess should be less than zero degrees C. i tried summation of Q = 0 but then i found out that I have no M (mass of ice) and T (final temperature)

• Answer:

  You will need a thermal equilibrium for the following situation: The heat needed to bring all the ice from [math]-30^\circ[/math] C to [math]0^\circ[/math] C is taken from the water and calorimeter. When 0 degree is reached for the ice, the remaining heat will start to melt the ice, of which you would like only half to be melted. Taking the heat transport equation: [math]Q_{down} = Q_{up}[/math], where [math]Q_{up}[/math] means the amount of heat needed to move the temperature of the subsystem up (Ice) and [math]Q_{down}[/math] is the amount of heat extracted from the subsystem (Calorimeter + Water) to deliver said heat. Filling in all parameters we get: [math]M_{cal} C_{cal} \Delta T_{cal} + M_{wat} C_{wat} \Delta T_{wat} = [/math] [math]M_{ice} C_{ice} \Delta T_{ice} + \frac{1}{2} M_{ice} C_{ice,melt}[/math], with [math]M[/math] the mass, [math]C[/math] the heat capacity and [math]\Delta T[/math] the temperature change. Except for [math]M_{ice}[/math], we know the values for all the variables (see exercise for the heat capacity numbers of the calorimeter, water, ice and melting ice). The temperature differences are as follows: [math]\Delta T_{wat} = \Delta T_{cal} = 25^\circ[/math] C and [math]\Delta T_{ice} = 30^\circ[/math] C. Using the following additional values ([math]M_{wat}[/math] is unknown): [math]C_{wat} = 4186[/math] J/kgK [math]C_{ice} = 2050[/math] J/kgK [math]C_{ice,melt} = 334[/math] J/kg Leaves us with the following equation: [math]M_{ice} = 3.65 + 1.70 M_{wat}[/math]