Đề thi thử Tốt nghiệp THPT môn Vật lí THPT Hoàng Văn Thụ (Nam Định) có đáp án

${m_1}{c_1}\left( {{t_1} - t} \right) = {m_2}\left( {\lambda + {c_2}t} \right) \Rightarrow {m_1} \cdot 4000 \cdot (85 - 40) = {m_2}\left( {334 \cdot {{10}^3} + 4200 \cdot 40} \right) \Rightarrow {m_2} = \frac{{90}}{{251}}{m_1}$

$\frac{{{C_1} - {C_2}}}{{{C_1}}} = 1 - \frac{{{C_2}}}{{{C_1}}} = 1 - \frac{{\frac{{{m_{cp}}}}{{{m_1} + {m_2}}}}}{{\frac{{{m_{cp}}}}{{{m_1}}}}} = 1 - \frac{{{m_1}}}{{{m_1} + {m_2}}} = 1 - \frac{1}{{1 + \frac{{90}}{{251}}}} \approx 0{,}264 = 26{,}4\% {\rm{. }}$Chọn B

$n = \frac{m}{M} = \frac{{0{,}5 \cdot {{10}^3}}}{{235}} = \frac{{100}}{{47}}\;{\rm{mol}}$

$N = n{N_A} = \frac{{100}}{{47}} \cdot 6{,}02 \cdot {10^{23}} \approx 1,{281.10^{24}}$

$Q = N\Delta E = 1{,}281 \cdot {10^{24}} \cdot 200 \cdot 1{,}6 \cdot {10^{ - 13}} \approx 4{,}1 \cdot {10^{13}}J$

$A = HQ = 0{,}2.4,{1.10^{13}} = 8,{2.10^{12}}J$

$t = \frac{Q}{P} = \frac{{8,{{2.10}^{12}}}}{{{{160.10}^3}}} = {5125.10^4}s \approx 593$ ngày. Chọn B

$\left\{ {\begin{array}{llllllllllllllllllll}{{N_X} = {N_0} \cdot {2^{\frac{{ - t}}{T}}}}\\{{N_Y} = {N_{0Y}} + {N_0} \cdot \left( {1 - {2^{\frac{{ - t}}{T}}}} \right) \Rightarrow \left\{ {\begin{array}{llllllllllllllllllll}{0{,}75 = {2^{\frac{{ - {t_1}}}{T}}}}\\{0{,}5{N_0} = {N_{0Y}} + {N_0} \cdot \left( {1 - {2^{\frac{{ - t}}{T}}}} \right)}\end{array}} \right.}\\{ \Rightarrow 0{,}5{N_0} = {N_{0Y}} + {N_0} \cdot (1 - 0{,}75) \Rightarrow {N_{0Y}} = 0{,}25{N_0}}\end{array}} \right.$

Chọn B

${m_n} = VD = 1{,}5.1 = 1{,}5\;{\rm{kg}}$

$Q = \left( {{m_n}{c_n} + {m_{nh}}{c_{nh}}} \right)\Delta t + 0{,}3{m_n}L = (1{,}5 \cdot 4190 + 0{,}6 \cdot 880) \cdot (100 - 20) + 0{,}3 \cdot 1{,}5 \cdot 2{,}26 \cdot {10^6} = 1562040J$

$A = \frac{Q}{H} = \frac{{1562040}}{{0{,}75}} = 2082720\;{\rm{J}}$

$P = \frac{A}{t} = \frac{{2082720}}{{40.60}} = 867{,}8\;{\rm{W}} \Rightarrow$ a) Đúng; b) Sai; c) Sai; d) Đúng

${p_1} = {p_0} + \frac{{{m_{gh}}g}}{S} = {10^5} + \frac{{5 \cdot 10}}{{50 \cdot {{10}^{ - 4}}}} = 1{,}1 \cdot {10^5}\;{\rm{Pa}} \Rightarrow$ a) Sai

${p_2} = {p_0} + \frac{{\left( {{m_{hs}} + {m_{gh}}} \right)g}}{S} = {10^5} + \frac{{(55 + 5) \cdot 10}}{{50 \cdot {{10}^{ - 4}}}} = 2{,}2 \cdot {10^5}Pa \Rightarrow$ b) Đúng

Quá trình ghế hạ xuống, khí trong xilanh nhận công $\Rightarrow$ c) Sai

$pV =$ const $\Rightarrow {p_1} \cdot Sl = {p_2} \cdot S(l - 8) \Rightarrow 1{,}1 \cdot {10^5}.l = 2{,}2 \cdot {10^5}.(l - 8) \Rightarrow l = 16\;{\rm{cm}} \Rightarrow$ d) Đúng

a) Đúng

$\frac{{{p_1}{V_1}}}{{{T_1}}} = nR \Rightarrow \frac{{8{,}31 \cdot {{10}^5} \cdot {V_1}}}{{27 + 273}} = 8{,}31 \Rightarrow {V_1} = 3 \cdot {10^{ - 3}}\;{{\rm{m}}^3}$

${T_1} = aV_1^2 \Rightarrow 27 + 273 = a \cdot {\left( {3 \cdot {{10}^{ - 3}}} \right)^2} \Rightarrow a = \frac{{10}}{3} \cdot {10^7}\left( {\;{\rm{K}}/{{\rm{m}}^6}} \right) \Rightarrow$b) Sai

$\frac{{{p_2}{V_2}}}{{{T_2}}} = nR \Rightarrow \frac{{13{,}85 \cdot {{10}^5} \cdot {V_2}}}{{\frac{{10}}{3} \cdot {{10}^7} \cdot V_2^2}} = 8{,}31 \Rightarrow {V_2} = 5 \cdot {10^{ - 3}}\;{{\rm{m}}^3} \Rightarrow {T_2} = 833{,}3\;{\rm{K}} \Rightarrow$c) Đúng

Từ 1 đến 2 có $nR = \frac{{pV}}{T} = \frac{{pV}}{{a{V^2}}} = \frac{p}{{aV}} \Rightarrow p = nRaV \Rightarrow {\rm{p}}$ theo V là hàm bậc nhất

$A_{12}^\prime = \frac{1}{2}\left( {{p_1} + {p_2}} \right)\left( {{V_2} - {V_1}} \right) = \frac{1}{2}\left( {8{,}31 \cdot {{10}^5} + 13{,}85 \cdot {{10}^5}} \right)(5 - 3) \cdot {10^{ - 3}} = 2216J \Rightarrow$d) Đúng

a) Đúng

b) Đúng. Sau 1T thì một nửa số hạt Na hiện có bị phân rã.

c) Sai. ${n_0} = \frac{m}{M} = \frac{{0{,}248 \cdot {{10}^{ - 3}}}}{{25}} = 9{,}92 \cdot {10^{ - 6}}\;{\rm{mol}}$

${N_0} = {n_0}{N_A} = 9{,}92 \cdot {10^{ - 6}} \cdot 6{,}02 \cdot {10^{23}} \approx 5{,}972 \cdot {10^{18}}$

$H = \lambda N = \frac{{\ln 2}}{T} \cdot {N_0} \cdot {2^{\frac{{ - t}}{T}}} = \frac{{\ln 2}}{{62}} \cdot 5{,}972 \cdot {10^{18}} \cdot {2^{\frac{{ - 10.60}}{{62}}}} \approx 8{,}15 \cdot {10^{15}}\;{\rm{Bq}}$

d) Đúng. $\frac{H}{{{H_0}}} = {2^{\frac{{ - t}}{T}}} = {2^{\frac{{ - 186}}{{62}}}} = 0{,}125 = 12{,}5\%$

$T = 5\hat o = 5.0{,}004 = 0{,}02s \Rightarrow f = \frac{1}{T} = 50\;{\rm{Hz}}$

Trả lời ngắn: 50

${I_0} = 5\hat o = 5A \Rightarrow I = \frac{{{I_0}}}{{\sqrt 2 }} \approx 3{,}5A$

Trả lời ngắn: 3,5

$S = \pi {r^2} + 2\pi rh = \pi \cdot {4^2} + 2\pi \cdot 4 \cdot 10 = 96\pi \left( {\;{\rm{c}}{{\rm{m}}^2}} \right)$

$V = \pi {r^2}h = \pi \cdot {4^2} \cdot 10 = 160\pi \left( {\;{\rm{c}}{{\rm{m}}^3}} \right)$

$m = VD = 160\pi \cdot {10^{ - 6}} \cdot 1000 = 0{,}16\pi (\;{\rm{kg}})$

$- mc \cdot d{t_t} = \frac{{kS\left( {{t_t} - {t_p}} \right)}}{d} \cdot d\tau \Rightarrow - \frac{1}{{{t_t} - {t_p}}} \cdot d{t_t} = \frac{{kS}}{{mcd}} \cdot d\tau \Rightarrow - \int_{100}^4 {\frac{1}{{{t_t} - 25}}} \cdot d{t_t} = \int_0^2 {\frac{{96\pi \cdot {{10}^{ - 4}}}}{{0{,}16\pi \cdot 4180 \cdot 5 \cdot {{10}^{ - 3}}}}} \cdot d\tau$

Casio shift solve $\Rightarrow {t_t} \approx 99,{5706^\circ }{\rm{C}}$

$Q = mc\Delta T = 0{,}16\pi \cdot 4180.(100 - 99{,}5706) \approx 902J$

Trả lời ngắn: 902

$- mc \cdot d{t_t} = \frac{{kS\left( {{t_t} - {t_p}} \right)}}{d} \cdot d\tau \Rightarrow - \frac{1}{{{t_t} - {t_p}}} \cdot d{t_t} = \frac{{kS}}{{mcd}} \cdot d\tau \Rightarrow - \int_{100}^t {\frac{1}{{{t_t} - 25}}} \cdot d{t_t} = \int_0^{30} {\frac{{96\pi \cdot {{10}^{ - 4}}}}{{0{,}16\pi \cdot 4180 \cdot 5 \cdot {{10}^{ - 3}}}}} \cdot d\tau$

Casio shift solve $\Rightarrow {t_t} \approx 93,{8^\circ }{\rm{C}} \Rightarrow \Delta {t_t} = 100 - 93{,}8 = 6,{2^\circ }{\rm{C}}$

Trả lời ngắn: 6,2

$A = 210$

Trả lời ngắn: 210

${n_0} = \frac{{{m_0}}}{M} = \frac{{100}}{{210}} = \frac{{10}}{{21}}mol$

$${N_0} = {n_0}{N_A} = \frac{{10}}{{21}} \cdot 6{,}02 \cdot {10^{23}} \approx 2{,}87 \cdot {10^{23}}$$

$$N = {N_0} \cdot {2^{\frac{{ - t}}{T}}} = 2{,}87 \cdot {10^{23}} \cdot {2^{\frac{{ - 69}}{{138}}}} \approx 20 \cdot {10^{22}}$$

Trả lời ngắn: 20