All categories

4 years ago

How many "big c" calories does it take to raise the temperature of 2.00 l of water from 22.0 c to 40.0 c?

Physics

1 answer:

nignag [31]4 years ago

7 0

given that

mass of water = 2 L = 2000 gram

now we can use

heat required to raise the temperature

$Q = mC\Delta T$

C = 1 cal/g C

$\Delta T = 40 - 22 = 18$

now from above formula

$Q = 2000* 1 * 18$

$Q = 36000 Cal$

so above is the heat required to raise the temperature

You might be interested in

A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the ma

masha68 [24]

Answer:

$v=21.36\,\,\frac{m}{s}\\$

$m=1.2357\,\,kg$

Explanation:

Recall the formula for linear momentum (p):

$p = m\,v$ which in our case equals 26.4 kg m/s

and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:

$K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}$

Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:

$K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg$

Now by knowing the particle's mass, we use the momentum formula to find its speed:

$p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}$

4 0

3 years ago

Earth is slightly closer to the Sun in January than in July. How does the area swept out by Earth's orbit around the Sun during

suter [353]

Answer: Option a.

Explanation:

Kepler's 2nd law of planetary motion states:

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.

3 0

3 years ago

You place a piece of aluminum at 250.0∘C ∘ C in 9.00 kg k g of liquid water at 20.0∘C ∘ C . None of the water boils, and the fin

Mice21 [21]

Answer:

m₁ = 0.37 kg

Explanation:

According to Law of conservation of energy:

Heat Lost by Aluminum = Heat Gained by Water

m₁C₁ΔT₁ = m₂C₂ΔT₂

where,

m₁ = mass of piece of aluminum = ?

C₁ = specific heat capacity of aluminum = 900 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 250°C - 22°C = 228°C

m₂ = mass of water = 9 kg

C₂ = specific heat capacity of water = 4200 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 22°C - 20°C = 2°C

Therefore,

m₁(900 J/kg.°C)(228 °C) = (9 kg)(4200 J/kg.°C)(2°C)

m₁ = (75600 J)/(205200 J/kg)

m₁ = 0.37 kg

5 0

3 years ago

ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements. You travel 20.0 mph f

timurjin [86]

The actual answer is 165 miles, but using significant figure rules the answer is 200. This is because the sig fig rules are as follows ...
1. Non-zero digits are always significant.
2. Any zeros between two significant digits are significant.
3. A final zero or trailing zeros in the decimal portion ONLY are significant.

So the zeroes in a number like 20 or 23,000 are NOT significant. When you add numbers you must find the addend with the lowest amount of significant figures and round the answer to that. In this case most of the addends only have one sig fig, so you round 165 to 200 to make it only have one sig fig.

5 0

4 years ago

Read 2 more answers

A woman standing on the ground sees a rocket ship move past her at 95% the speed of light. Compared to the rocket at rest, the w

Rufina [12.5K]

As per Einstein's theory of relativity we know that when an object will move with the speed comparable to the speed of light then the length of the object will be different from its length at rest position

This is also known as length contraction theory

As we know here that

$L = L_0\sqrt{1 - \frac{v^2}{c^2}}$

so here we know that

v = 0.95 c

so from above equation we will have

$L = L_0\sqrt{1 - 0.95^2}$

$L = 0.31 L_0$

so here the length will be SHORTER

8 0

3 years ago