Question #78236

The specific heat of plastic is 50 times greater than the specific heat of lead, 10 times greater than the specific heat of stone, and 1.5 times greater than the specific heat of water. If the samples of plastic, lead, stone, and water have identical masses and identical starting temperatures, and are given identical increases in energy of 1,000 J, which sample will end up with the highest temperature?

A. plastic
B. lead
C. stone
D. water
E. None of the Above
1

Expert's answer

2018-06-19T05:26:08-0400

Question #78236, Chemistry / General Chemistry

The specific heat of plastic is 50 times greater than the specific heat of lead, 10 times greater than the specific heat of stone, and 1.5 times greater than the specific heat of water. If the samples of plastic, lead, stone, and water have identical masses and identical starting temperatures, and are given identical increases in energy of 1,000 J, which sample will end up with the highest temperature?

A. plastic

B. lead

C. stone

D. water

E. None of the Above

SOLUTION

Q=cmΔt,Q = c * m * \Delta t,


in it Q is the heat energy, J;

c1c_1 is the specific heat of plastic J/(g+C)J/(g^{+*}C);

c2c_2 is the specific heat of lead J/(g+leadC)J/(g^+ \text{lead}^*C);

c3c_3 is the specific heat of stone J/(g+C)J/(g^{+*}C);

c4c_4 is the specific heat of water J/(g+C)J/(g^{+*}C);

m is the mass of the substance, g;

Δt1\Delta t_1 is the change temperature of plastic in C^\circ \mathrm{C};

Δt2\Delta t_2 is the change temperature of lead in C^\circ \mathrm{C};

Δt3\Delta t_3 is the change temperature of stone in C^\circ \mathrm{C};

Δt4\Delta t_4 is the change temperature of water in C^\circ \mathrm{C};


Δt1=Q/(c1m),Δt2=Q/(c2m),Δt3=Q/(c3m),Δt4=Q/(c4m),\begin{array}{l} \Delta t_1 = Q / (c_1 * m), \\ \Delta t_2 = Q / (c_2 * m), \\ \Delta t_3 = Q / (c_3 * m), \\ \Delta t_4 = Q / (c_4 * m), \end{array}Δt1:Δt2:Δt3:Δt4=1,000c1m:501,000c1m:101,000c1m:1.51,000c1m=1:50:10:1.5,\Delta t_1: \Delta t_2: \Delta t_3: \Delta t_4 = \frac{1,000}{c_1 * m}: \frac{50 * 1,000}{c_1 * m}: \frac{10 * 1,000}{c_1 * m}: \frac{1.5 * 1,000}{c_1 * m} = 1:50:10:1.5,Δt2>Δt3>Δt4>Δt1\Delta t_2 > \Delta t_3 > \Delta t_4 > \Delta t_1


the change temperature of lead is 50 times greater of plastic;

the change temperature of stone is 10 times greater of plastic;

the change temperature of water is 1.5 times greater of plastic.


Δt=tendtstarttend=tstart+Δt,tstart1=tstart2=tstart3=tstart4\begin{array}{l} \Delta t = t_{\text{end}} - t_{\text{start}} \\ t_{\text{end}} = t_{\text{start}} + \Delta t, \\ t_{\text{start1}} = t_{\text{start2}} = t_{\text{start3}} = t_{\text{start4}} \\ \end{array}


then t2end>t3end>t4end>t1endt_{2\text{end}} > t_{3\text{end}} > t_{4\text{end}} > t_{1\text{end}}

ANSWER:B.

Answer provided by AssignmentExpert.com


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS