Tutor profile: William B.

A block of iron weighing 3.60 kg at a temperature of 807.0 °C was inserted into a container containing 1.00 liter of water at a temperature of 30.0 °C isolated from the environment. The cooling process of iron happens in two phases, the first phase the water is heated to the boiling point and in the second stage the water evaporates. This phase continues until the temperature of the iron is equal to the temperature of the water. The final temperature of the water and the iron is 100.0 °C. How many liters of liquid water will remain at the end of the process? $(\text{The specific heat of solid iron is} \ 0.450 J g^{¯1 }K^{¯1}\ \text{ and the heat of vaporization for water is} \ 2259.23 \frac{J}{g}.$)

We know the change in the temperature of the iron. $(\Delta T_{Fe}=807-100=707.$) Now, using the following formula for heat we can find the energy for the change in iron: $(Q=mc\Delta T.$) That is, $(Q_{Fe}=m_{Fe}\cdot c_{Fe}\cdot \Delta T_{Fe}=(3600 g) (707.0 K) (0.450 J g^{¯1} K^{¯1})=1145340 J. $) Now, using the same idea to see how much energy is required to raise the temperature of the water. $(Q_{H_2O}=m_{H_2O}\cdot c_{H_2O}\cdot \Delta T_{H_2O}=(1000 g) (70.0 C) (4.184 J g^{¯1} K^{¯1})=292880 J$) Note: Kelvin and Celsius move at the same rate and 1.00L=1000g, so we can make the unit conversions. Furthermore, $(Q_{system}=Q_{Fe}-Q_{H_2O}=1145340J-292880 J=852460 J$) At this point, we need to calculate the boil off or how much water is evaporated. We can do this by: $(H_{vap}=\frac{Q_{system}}{m_{object}}.$) Rearranging gives $(m_{H_2O}=\frac{Q_{system}}{H_{vap}}=\frac{852460 J}{\ 2259.23 \frac{J}{g}}=377.3 g.$) Finally, using the idea of converting grams to liters we can see that $(m_{H_2O}=m_{int}-m_{vap}=1000g-377.3g\approx 623g=.623L$)

Find the derivative of the following: $(y=\ln(x\cdot \sin(x^2)$)

This is a chain rule problem involving the $(\ln(x\cdot \sin(x^2)).$) Recall the chain rule as the following: $(\text{If} \ y=f(u(x)) \ \text{(A function composed with another function), then} \frac{dy}{dx}=\frac{du}{dx}\cdot \frac{dy}{du}.$) We need to find the following derivatives $(\frac{dy}{du}, \ \text{where} \ y=\ln(u), \ \text{and} \ \frac{du}{dx} , \ \text{where} \ u=x\cdot \sin(x^2).$) Thus, $(\frac{dy}{du} =\frac{1}{u}. \ \text{Converting back to the x coordinate gives} \ \frac{dy}{du} =\frac{1}{x\cdot \sin(x^2)}. $) Now, $(\frac{du}{dx} \ \text{is more complicated because we need to use the product rule.} $) Therefore, we notice that $(u=f(x)\cdot g(x), \ \text{where} \ f(x) =x \ \text{and} \ g(x)=\sin(x^2).$) Note: The product rule is as follows: $(\text{If} \ u=f(x)\cdot g(x), \ \text{then} \ \frac{du}{dx}=f'(x)\cdot g(x)+g'(x)\cdot f(x).$) Therefore, $(f'(x)=1 \\ \text{and} \ g'(x)=2x \cos(x^2) \ \text{by the chain rule again, and using the derivative property of} \ \sin(x).$) Thus, $(\frac{du}{dx}=\sin(x^2)+2x^2\cdot \cos(x^2).$) Finally, putting everything together we get $(\frac{dy}{dx}=\frac{du}{dx}\cdot \frac{dy}{du}=\frac{\sin(x^2)+2x^2 \cos(x^2)}{x\sin(x^2)}$)

A force F at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is µ. The work done by the frictional force is?

Since $(\Delta v=0$) implies that $(a=0, \text{and} \ F=ma=0.$) Thus, $(F_{\mu}+F_x=0 \implies F_\mu=-F_x$) Moreover, we need to find the horizontal force component. We will use right angle trigonometry to find the follow: $( F_x=F\cdot\cos(\theta).$) Now, $(W_\mu=F_\mu \cdot d, \ \text{where d is the distance. } $) Thus, $(W_\mu=-F\cdot d\cdot \cos(\theta).$)