Q= mcΔt
50000. J = (5000g)(4.18 J/g°C)(Δt)
Δt= final temp-initial temp
Δt= 2.39°C
2.39°C= final temp-12.0°C
Final temp= 14.39°C
Explanation:
please can you send me the pic of Faith because I am not able to understand any of your location is too bad I can't answer sorry for stamp in which class you are in
An incandescent bulb becomes hotter than a fluorescent bulb when turned on because in a regular incandescent bulb, there is tungsten wire where electricity is converts into heat. A regular incandescent light bulb requires 4 times more energy than a fluorescent bulb in order to produce the same amount of light. The conversion is such that for a 75-watt bulb, temperature get raised to approximately 2000 K. For such a high temperature, the radiating energy from the wire have some visible light. In such bulbs, 90% of the electricity get consumed in producing heat and only 10% produces light thus, they are not much efficient source of light.
On the other hand, fluorescent bulbs produce light with less amount of heat. In them, 40% of electricity is consumed in producing light and 60% in heat which is very less as compared to heat produced by a incandescent bulb. This is because when it get turned on, mercury atoms inside the bulb collides with electrons and produce UV light which is then converted into visible light using thin layer of phosphor power present inside the bulb. This produces low amount of heat thus, the bulb stays cooler, the bigger size of bulb also helps in dispersing heat.
Therefore, a fluorescent light bulb is not as hot as an incandescent light bulb.
Answer: a, b, d
Explanation:Physical property are measurable (or perceived) property of something observable without having to change the composition or identity of that thing.
Examples of physical properties in the context included in the question are the following: • Temperature • Solubility • Resistivity • Conductivity • Density
Sn + CuSO4 → Cu + SnSO4 I think this should be the answer. Hope it helps.
