Answer:
24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)
Explanation:
24) This problem can be solved by means of the following equation.

where:
DU = internal energy difference [J]
Q = Heat transfer [J]
W = work [J]
Since there are no temperature changes the internal energy change is equal to zero
DU = 0
therefore:

The work is equal to the heat transfered, W = 75 [J].
25) The heat transfer can be calculated by means of the following equation.
![Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]](https://tex.z-dn.net/?f=Q%20%3D%20m%2Ac_%7Bp%7D%2ADT%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%200.4%5Bkg%5D%5C%5Cc_%7Bp%7D%20%3D%20specific%20heat%20%3D%20897%5BJ%2Fkg%2AK%5D%5C%5CDT%3D%205%20%5BC%5D)
Q = 0.4*897*5 = 1794[J]
Work is equal to heat transfer, W = 1794[J]
26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

where:
m = mass = 0.5[kg]
g = gravity = 9.81[m/s^2]
h = 1.5 [m]
![E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D0.5%2A9.81%2A1.5%5C%5CE_%7Bp%7D%3D7.36%5BJ%5D)
The heat developed can be calculated by means of the following equation.
![Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]](https://tex.z-dn.net/?f=Q%3Dm%2Ac_%7Bp%7D%2ADT%5C%5CQ%3D0.5%2A130%2A1%5C%5CQ%3D65%5BJ%5D)
The number of times will be calculated as follows
n = 65/7.36
n = 8.8 (times) or 9 (times)
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The coefficient of expansion is 13 * 10^-6 m per meter length.per oK
The temperature difference = 42 - - 8 = 50 oC
delta T = (42 + 273) - (-8 + 273) = 50 oK
delta L = L * 13* 10^6 m/oK
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m
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Answer:

Explanation:
First of all, we need to calculate the total energy supplied to the calorimeter.
We know that:
V = 3.6 V is the voltage applied
I = 2.6 A is the current
So, the power delivered is

Then, this power is delivered for a time of
t = 350 s
Therefore, the energy supplied is

Finally, the change in temperature of an object is related to the energy supplied by

where in this problem:
E = 3276 J is the energy supplied
C is the heat capacity of the object
is the change in temperature
Solving for C, we find:
