A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.
<h3>What is specific heat?</h3>
It is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).
A metallic rod of mass 150 g (m) absorbs 82.5 cal of heat (Q) and its temperature raises from 20 °C to 25 °C. We can calculate the specific heat (c) of the metal using the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 82.5 cal / 150 g × (25 °C - 20 °C) = 0.11 cal/g.°C
A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.
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Answer:
The impuise is 7.9905 kg*m/s
Explanation:
Step one:
given data
v1= +2.63m/s
v2=-20.2m/s
mass m= 0.350kg
Step two:
From the expression for impulse
Ft= mΔv
substituting our data into the expression we have
Ft= 0.35*(-20.2-2.63)
Ft= 0.35*22.83
Ft=7.9905 kg*m/s
Answer: Option B
Explanation:
Rocks contains pores and spaces in them in which water gets trapped, and this causes the rock to break as water increases its volume due to freezing. This rocks are then eroded and are carried away into the streams and channels. The rivers and streams can carry particles of various shapes and size, which are further transported and deposited at a different place.
This particles or sediments then gets compacted and lithified in due course of time, and forms sedimentary rocks.
Water is also needed to undergo metasomatism process, that creates a metamorphic rock.
Hence, water flow is also an important driving force in rock cycle, along with tectonic activities.
Thus, the correct answer is option (B).
Multiply the power (1,800 watts) by time (1,200 seconds) to get 2,160,000 joules (or 2.16 MJ)
Efficiency of heat engine is determined by the ratio of difference in temperature of cold from hot reservoir to the temperature of hot reservoir over temperature of hot reservoir.
Answer: Option A
<u>Explanation:</u>
Efficiency is defined as the output from the input. So it is the ratio of energy output to the energy input. In case of temperature, it is the change in temperature from hot reservoir to cold reservoir with the overall hot reservoir temperature.
Thus, option A is the most suitable as the heat will be transferred from high temperature to low temperature. So the hot reservoir will be releasing the energy. So the conversion of hot reservoir temperature to cool reservoir temperature is defined as the efficiency. Thus, option A is the most suitable.
