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zvonat [6]
3 years ago
7

It requires 350 joules to raise a certain amount of a substance from 10.0°C to 30.0°C. The specific heat of the substance is 1.2

J/g°C.
Physics
2 answers:
Akimi4 [234]3 years ago
6 0
I'll assume you are looking for the mass of the object, since that is the missing piece of the puzzle.

The important equation for heat and energy is

Energy = mass × specific heat × change in temperature

Things we know:
Energy needed is 350 J.
Specific heat = 1.2 J/g°C
Temp. change = (30-20)°

Now we just need to plug those in and rearrange the formula to find the mass!

350 = mass × 1.2 × 10
Ivanshal [37]3 years ago
6 0

Answer:

The amount of the substance is 14.6 g.

Explanation:

Given that,

Energy = 350 J

Temperature T_{1}=10.0^{\circ}\ C

Temperature T_{2}=30.0^{\circ}\ C

Specific heat capacity = 1.2 J/g^{/circ}[/tex]

We know that,

The formula of specific heat is defined as

Q = mc\Delta T

Where,

m = mass of the substance

c = specific heat capacity

\Delta T = Temperature change

Put the value into the formula

350=m\times 1.2\times(30-10)

The amount of substance is

m =\dfrac{350}{1.2\times(30-10)}

m =14.6\ g

Hence, The amount of the substance is 14.6 g.

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Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

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