A cube of ice is take from the freezer at -8.5C and placed in a 95g calorimeter aluminum cup containing 310g of water at 20degC. The final situation is observed to be all water at 17degC. What was the mass of the ice cube?

19.276g

m cice �

0 − (−8.5)�
| {z }
ice from −8.5
◦C to 0◦C
+ m Lwater
| {z }
ice melts at 0◦C
+ m cwater
17 − 0

| {z }
water from 0◦C to 17◦C
= 0.300cwater
20 − 17�
| {z }
water from 20◦C to 17◦C
+ 0.075cAl
20 − 17�
| {z }
Al from 20◦C to 17◦C
⇒ m

2090�
8.5 + m

3.33 × 105

+ m

4186�
17 = 0.300 �
4186�
3 + 0.075 �
900�
3
⇒ m

421,927�
= 3969.9

To solve this problem, we'll use the principle of conservation of energy and the specific heat capacity formula.

The principle of conservation of energy states that the energy gained or lost by a system must be equal to the energy lost or gained by the surroundings. In this case, the energy lost by the ice cube will be gained by the water in the calorimeter cup.

The specific heat capacity (C) is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius per unit mass. The specific heat capacity of water is approximately 4.18 J/g°C, and the specific heat capacity of aluminum is approximately 0.897 J/g°C.

To calculate the mass of the ice cube, we can use the formula:

Energy gained by water = Energy lost by ice cube

The energy gained by the water can be calculated using the formula:

Energy gained by water = (mass of water) x (specific heat capacity of water) x (change in temperature)

The energy lost by the ice cube can be calculated using the formula:

Energy lost by ice cube = (mass of ice cube) x (specific heat capacity of ice) x (change in temperature)

Since the final situation is observed to be all water, the energy lost by the ice cube will equal the energy gained by the water. Therefore, we can set up the equation:

(mass of water) x (specific heat capacity of water) x (change in temperature) = (mass of ice cube) x (specific heat capacity of ice) x (change in temperature)

Now let's plug in the values given in the problem:

mass of water = 310 g
specific heat capacity of water = 4.18 J/g°C
change in temperature of water = 17°C - 20°C = -3°C (note: we use the difference in temperature)

We also know that the specific heat capacity of ice is approximately the same as water (4.18 J/g°C).

Now we can solve for the mass of the ice cube:

mass of ice cube = (mass of water) x (specific heat capacity of water) x (change in temperature) / (specific heat capacity of ice) x (change in temperature)

mass of ice cube = (310 g) x (4.18 J/g°C) x (-3°C) / (4.18 J/g°C) x (-3°C)

Simplifying the equation:

mass of ice cube = 310 g

Therefore, the mass of the ice cube is 310 grams.

The sum of heats gained is zero.

HeatgainedbyicewarmigtoOC + heatgainedmeltingIcd+ heat gained by aluminum+heat gained by original water=0

massice*Cice*(0-8.5C)+massice*Hf+95*Calum*(17-20)+310*Cwater*(17-20)=0
solve for mass of ice.