Ask question

Ask question

All categories

2 years ago

5

They want a mixture of 150 grams of water at 60 degrees Celsius, using hot water at 90 degrees Celsius and cold water at 14 degr

ees Celsius, how many grams of water should be used per face?

1 answer:

Gnesinka [82]2 years ago

6 0

Answer:

Explanation:

Here,

Final mass(m)=150g

Final Temperature(t)=60°C

Temperature of hot water(t1)=90C

Temperature of cold water(t2)=14C

Let the grams of water to be used per face be m1 grams and (150-m1) grams for hot and cold water respectively.

We know that,

Heat lost by hot body=Heat gained by cold body

or, m1 s1 Δt1=m2 s2 Δt2

or, m1 (t1-t)=(150-m1) (t-t2) [∵s1=s2, SHC of water is taken to be same]

or, m1 (90-60)=(150-m1) (60-14)

or, 30m1=6900-46m1

or, 76m1=6900

∴m1=90.79g

(150-m1)=59.21g

Hence, 90.79g and 59.21 g hot and cold water must be used respectively.

You might be interested in

The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving ele

siniylev [52]

The valence electrons of metals are weakly attracted to the parent nuclei, so the electrons break free and float. The moving electrons form a electron <u>negative</u> blanket that binds the atomic <u>positive</u> nuclei together, forming a metallic bond.

So the answers are <u>{ Negative }</u> and <u>{ Positive }.</u>

Please vote Brainliest (:

5 0

2 years ago

Read 2 more answers

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

2 years ago

The French high-speed train travels at 300 km/h. How long

Agata [3.3K]

Answer:

given , v = 300 km/hr; distance d = 1500 km; then time t = d/v = 1500/300 = 5 hrs

Explanation:

4 0

3 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

$v_{1i}$ = initial velocity of the first body before collision = $v$

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

$v_{1f}$ = final velocity of the first body after collision =

using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

8 0

2 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

so

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago