Ask question

Ask question

All categories

3 years ago

8

180 cm3 of hot tea at 97 °C are poured into a very thin paper cup with 20 g of crushed ice at 0 °C. Calculate the final temperat

ure of the "ice tea."(Hint: think about two processes: melting the ice into liquid and (maybe) warming the melted ice—now liquid.)

1 answer:

Zolol [24]3 years ago

8 0

Answer : The final temperature of the mixture is $91.9^oC$

Explanation :

First we have to calculate the mass of water.

Mass = Density × Volume

Density of water = 1.00 g/mL

Mass = 1.00 g/mL × 180 cm³ = 180 g

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of hot water (liquid) = $4.18J/g^oC$

$c_2$ = specific heat of ice (solid)= $2.10J/g^oC$

$m_1$ = mass of hot water = 180 g

$m_2$ = mass of ice = 20 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of hot water = $97^oC$

$T_2$ = initial temperature of ice = $0^oC$

Now put all the given values in the above formula, we get

$(180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC$

$T_f=91.9^oC$

Therefore, the final temperature of the mixture is $91.9^oC$

You might be interested in

Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long

kipiarov [429]

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

$\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}$

Here,

$\delta$ = Change in final length $(l_0)$ and the initial length $l$

PART A)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{1.02-1}{1}$

$\epsilon = 0.01961$

PART B)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{2-1.05}{2}$

$\epsilon = 0.475$

PART C)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{3.07-3}{3}$

$\epsilon = 0.0233$

Therefore the rank of this deformation would be B>C>A

7 0

3 years ago

Type the correct answer in the box. Round your answer to the nearest tenth. Dina has a mass of 50 kilograms and is waiting at th

katen-ka-za [31]

I guess once you input the numbers into the correct places into the equation it would look like this :
PE = 50 * 9.8 * 5.0
PE = 2,450

6 0

3 years ago

Read 2 more answers

A certain helium-neon laser pointer, emitting light with a wavelength of 632 nm, has a beam with an intensity of 825 W/m2 and a

BabaBlast [244]

Answer:

$N = 1.62 \times 10^{15}$

Explanation:

Energy of each photon is given as

$E = \frac{hc}{\lambda}$

here we will have

$\lambda = 632 nm$

now we will have

$E = \frac{(6.626\times 10^{-34})(3 \times 10^8)}{632 \times 10^{-9}}$

$E = 3.14 \times 10^{-19} J$

now let say there is N number of photons per second

so power due to photons is

$P = N(3.14 \times 10^{-19})$

now intensity is given as power received per unit area

so we have

$I = \frac{P}{A}$

$825 = \frac{N(3.14 \times 10^{-19})}{\pi (1.40 \times 10^{-3})^2}$

$825 = N(5.11 \times 10^{-14})$

$N = 1.62 \times 10^{15}$

5 0

3 years ago

When the image of a distant object is brought into focus in front of a person's retina, the defect is called

Stells [14]

A. Nearsightedness
[Source] google

3 0

3 years ago

Demonstrate that plants use photosynthesis to grow throughout their lives

gavmur [86]

By definition, photosynthesis is the process made by green plants, algae and some other forms of bacteria to make their own food by converting carbon dioxide and water into glucose.

So how important is photosynthesis to plants? It is as important as how humans need food in order to survive. Plants also need food and energy to survive and grow.

Not only food is the important matter to be addressed when it comes to photosynthesis. It is also important for plants' cellular respiration. These processes are constantly made by plants throughout their entire life.

3 0

3 years ago