Calculate the heat needed to raise the ice temperature:

$Q_{heat ice} = m c_{ice} Delta T$

$Q_{heat ice} = (0.100 Kg)*(2060 J/Kg.C)*(0 – (-20))$

$Q_{heat ice} = 4120 J$

Calculate the heat needed to melt ice:

$Q_{melt} = m H_f$

$Q_{melt} = (0.100)*(3.34e5)$

$Q_{melt} = 33400 J$

The total amount of heat is:

$$
Q = Q_{heat ice} + Q_{melt} = 4120 + 33400 = 3.75 times 10^4 J
$$

$$
3.75 times 10^4 J
$$

Calculate the heat needed to raise the water temperature:

$Q_{heat water} = m c_{water} Delta T$

$Q_{heat water} = (0.200 Kg)*(4180 J/Kg.C)*(100 – 60)$

$Q_{heat water} = 33440 J$

Calculate the heat needed to vaporize water:

$Q_{vaporization} = m H_v$

$Q_{vaporization} = (0.200)*(2.26e6)$

$Q_{vaporization} = 452000 J$

Calculate the heat needed to raise the vapor temperature:

$Q_{heat vapor} = m c_{vapor} Delta T$

$Q_{heat vapor} = (0.200 Kg)*(2020 J/Kg.C)*(140 – 100)$

$Q_{heat vapor} = 16160 J$

The total amount of heat is:

$$
Q = 33440 + 452000 + 16160 = 5.02 times 10^5 J
$$

$$
5.02 times 10^5 J
$$

Calculate the heat needed to raise the ice temperature:

$Q_{heat ice} = m c_{ice} Delta T$

$Q_{heat ice} = (0.300 Kg)*(2060 J/Kg.C)*((0.00) – (-30))$

$Q_{heat ice} = 18540 J$

Calculate the heat needed to melt ice:

$Q_{melt} = m H_f$

$Q_{melt} = (0.300)*(3.34e5)$

$Q_{melt} = 100200 J$

Calculate the heat needed to raise the water temperature:

$Q_{heat water} = m c_{water} Delta T$

$Q_{heat water} = (0.300 Kg)*(4180 J/Kg.C)*(100 – 0.00)$

$Q_{heat water} = 125400 J$

Calculate the heat needed to vaporize water:

$Q_{vaporization} = m H_v$

$Q_{vaporization} = (0.300)*(2.26e6)$

$Q_{vaporization} = 678000 J$

Calculate the heat needed to raise the vapor temperature:

$Q_{heat vapor} = m c_{vapor} Delta T$

$Q_{heat vapor} = (0.300 Kg)*(2020 J/Kg.C)*(130 – 100)$

$Q_{heat vapor} = 18180 J$

The total amount of heat is:

$$
Q = 18540 + 100200 + 125400 + 678000 + 18180 = 9.40 times 10^5 J
$$

$$
9.40 times 10^5 J
$$

The first law of thermodynamics:

$Delta U = Q – W$

The change in thermal energy of an object is equal to the heat added to the
object minus the work done by the object.

The temperature of the balloon is the same, therefore its thermal energy is the same: $Delta U = 0$.

$Delta U = 0 = Q – W$

The balloon absorbed 75 J of heat: $Q=75 J$:

$0 = Q – W$

$0 = 75 – W$

Thus:

$$
W = 75 J
$$

$$
75 J
$$

$W = Q = m c_{aluminum} Delta T$

$W = (0.40 Kg) (897 J/kg.C) (5.0 C)$

$$
W = 1800 J
$$

$$
1800 J
$$

The energy of the shot increases by “$m g h$” per drop. We must drop it N times to raise its energy by “$m c_{lead} Delta T$”.

$N m g h = m c_{lead} Delta T$

Solve for N:

$N = dfrac{c_{lead} Delta T}{g h} = dfrac{(130)*(1.0)}{(9.80)*(1.5)}$

$$
N = 8.8 drops
$$

$$
8.8 drops
$$

Firstly,we should use $Delta U = m C Delta T$ equation ;

$Delta U =(0.15 kg)(4180J/kg.text{textdegree} C)(2.0 text{textdegree} C)$
$Delta U = 1.3 times 10^{3} J$

For the number of stirs ;

$dfrac{1.3 times 10^{3} J }{0.050 J}=2.6 times 10^{4}$

$textit$$text{color{#c34632} 2.6 times 10^{4} $$}$

The first law of thermodynamics states that the change in internal energy $Delta U$ of the system is the sum of the heat added to the system $Q$ and the work done by the system $W$. In other words,
$$
Delta U=Q-W
$$
Since the internal energy of the system is proportional to the temperature of the system if we want to cool down the system we need to have less internal energy at the end of the process or in other words we need to have negative change in the internal energy $Delta U<0$.

Looking at the first law of thermodynamics we can see that this can be achieved in two simple cases:
$$ W=0,:Q0$$
The first case corresponds to the system which is not doing any work nor the work is done on it (volume does not change) but it transfers heat to some colder object that it is in contact with.

The second case corresponds to the system which does not exchange heat with its surroundings but it does some work against the forces acting on it (expands).