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OLga [1]
3 years ago
8

It is necessary to determine the specific heat of an unknown object. The mass of the object is 201.0 g. It is determined experim

entally that it takes 15.0 J to raise the temperature 10.0 ° C. What is the specific heat of the object?
Physics
2 answers:
navik [9.2K]3 years ago
6 0
Mass = 0.201kg
Energy = 15J
temperature change = 10C

Energy(E) = mass(m) × specific heat capacity(c) × temperature change(θ)

we can rearrange this to make specific heat capacity the subject

c =\frac{E}{m\theta}

c =\frac{15}{2.01}
c =7.46268657

aliina [53]3 years ago
5 0

Answer:

The specific heat of the object is 0.0074 J/(g °C)

Explanation:

Data

mass, m = 201 g

Heat, Q = 15 J

temperature change, ΔT = 10 °C

specific heat, cp = ?

The equation that relates these variables is:

Q = m*cp*ΔT

Solving for specific heat we get:

cp = Q/(m*ΔT)

Replacing with data (units are ommited):

cp = 15/(201*10)

cp = 0.0074 J/(g °C)

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A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
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Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

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