What is the specific heat of a material that gains 600 joules of energy when a 0.25 kg object increases in temperature by 3C?
1 answer:
You might be interested in
Answer:
305 °F
Step-by-step explanation:
The core temperature of the object after 4 hours can be found using an exponential decay formula to model the decay of the difference between core temperature and ambient.
<h3>Cooling Model</h3>The solution to the differential equation described by Newton's law of cooling is the exponential equation ...
y = ab^t +c
where 'a' is the initial core temperature difference from ambient, 'b' is the decay factor of that difference in 1 unit of time period t. 'c' is the ambient temperature.
For this problem, the ambient temperature is c=80, and the differences of interest are ...
a = 1200 -80 = 1120
b = (830 -80)/1120 = 75/112
Using these values in the model gives ...
y = 1120(75/112)^t +80 . . . . . . where y(t) is the core temperature at time t
Note that units of time are hours.
<h3>Application</h3>We want y when t=4.
y = 1120(75/112)^4 +80 ≈ 1120(0.20108) +80 ≈ 305.212
The core temperature after 4 hours is about 305 °F.
__
<em>Additional comment</em>
The differential equation will have a solution of the form ...
where k = ln(75/112) ≈ -0.40101
In the above, we defined b = e^k = 75/112. Accuracy with this fraction can be better than using a truncated value of k.
Answer:
1/4th
Step-by-step explanation:
This comes from knowing the unit circle in both radians and degrees. However, you could also calculate it from knowing that 360 degrees is 2 radians. What you'll find is that
radians is 90 degrees, or 1/4th of the circle. This means that an arc measuring
radians will be 1/4th of the entire circumference.
I hope this helps!