Question

A) You want to cool a substance, which is put into a refrigerator that is at...

A) You want to cool a substance, which is put into a refrigerator that is at a constant temperature of 5centigrates . After 30 minutes, the substance is at 8 centigrates  and after 40 minutes it is at 6 centigrates . Find the initial body temperature.

B)  A substance cools from 100 centigrates  to 75  in 15 minutes in the open air (air temperature 20 centigrates ), find the temperature after 30 minutes.

Math   Calculus   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Water is boiled and is put to cool. The temperature of the air in the room...
Water is boiled and is put to cool. The temperature of the air in the room is rising linearly in relation to the function T(t)=30+0.01t, where t is minutes and T is Celsius. Suppose this satisfies Newton's Law of Cooling: the change of temperature of the water is proportional to the difference of the water temperature and the room’s ambient temperature. We take the water's temperature 10 minutes after and find that it's 81*C. Graph and explain T. Establish and...
A body at an unknown temperature is placed in a room which is held at a...
A body at an unknown temperature is placed in a room which is held at a constant temperature of 30° F. If after 10 minutes the temperature of the body is 0° F and after 20 minutes the temperature of the body is 15° F, find the unknown initial temperature.  I need to use MATLAB for this. Thankyou! Finals Answers:-30° F
You want to cool 0.2 kg of coffee, initially at temperature Th = 80° C, with...
You want to cool 0.2 kg of coffee, initially at temperature Th = 80° C, with ice initially at Tc = 0° C. The specific heat of ice is about 2108 J/kg K, and its latent heat of melting is about 334, 000 J/kg. You may take the specific heats of liquid water and coffee to be the same: 4187 J/kg K. A) Assume the coffee and ice form a closed system. You want them to equilibrate at 40° C....
Suppose the rate of change of the temperature of a body is proportional to the difference...
Suppose the rate of change of the temperature of a body is proportional to the difference between & the temperature of the body and the temperature of the surrounding environment which is 35 . The k initial body temperature is 120 After 40 minutes, the temperature went to 60 . a.) Find the differential equation with initial condition that models this situation. b.) Solve to find the general solution using the technique of separation of variables. Call the constant C...
6. Where you are now, you want to have tea while solving questions. The water, which...
6. Where you are now, you want to have tea while solving questions. The water, which is found in a thin steel teapot with a diameter of 12 cm, is boiling. The heat transport coefficient between the teapot base and the boiling water is 10 kW/m2K. Since the heat flow ingrair is 40 kW/m2; a) Find the surface temperature of the teapot base. b) Find the heat given to the water for 30 minutes and the amount of water evaporated.
This question is about Newton’s law of cooling, which states that the temperature of a hot...
This question is about Newton’s law of cooling, which states that the temperature of a hot object decreases proportionally to the difference between its temperature and the temperature of the surroundings. This can be written as dT dt = −k(T − Ts), where T is the temperature, t is time, k is a constant and Ts is the temperature of the surroundings. For this question we will assume that the surroundings are at a constant 20◦ and A that the...
One advantage to being small is that it is easier to cool off after exertion than...
One advantage to being small is that it is easier to cool off after exertion than if you are large. For example, hummingbirds generate a lot of heat as their wings beat quickly, but they can cool off very quickly in response. (The flip side is, it is harder to stay warm when the air temperature is low.) In this problem we will develop a simple model of the smallest shrew and a polar bear cooling off. We will model...
At 1 p.m., a thermometer reading 80°F is taken outside where the temperature is constant. At...
At 1 p.m., a thermometer reading 80°F is taken outside where the temperature is constant. At 1:05 p.m., the thermometer reads 35° F more than the surrounding temperature. Forty minutes after being taken outside, the thermometer reading differs from the outside temperature by 15° F. Determine the outside temperature. How long will it take for the same thermometer to cool from 100° F to 60° F in air, which is maintained at 50° F? Ans.: 40.5° F, 66.5 min
Based on the put-call parity relationship you want to make an arbitrage profit by selling a...
Based on the put-call parity relationship you want to make an arbitrage profit by selling a call, buying a put, and taking a leveraged equity position. Stock proce = $100 Call price (6-month maturity with strike price of $110) = $5 Put price (6-month maturity with strike price of $110) = $8 Risk free interest rate (continuously compounded) = 10% If the stock price at maturity is $120, how much do you earn from all these positions?
1. An oven chimney is made of brick with a heat transmission coefficient 1.1 W/mK 10...
1. An oven chimney is made of brick with a heat transmission coefficient 1.1 W/mK 10 cm thick. Since the shaft exterior radiation beam coefficient is 0.8 and the flue gas temperature is 350°C, the external ambient temperature is 25°C and the outer ambient heat transport coefficient is 20 W/m2K; a) Calculate the shaft exterior temperature. b) In order to reduce the risk of burn injury that may occur in the body, the shaft is asked to be below 55°Cve...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT