Question

An aluminum block (c = 9.1 X 10² J/kg K) of mass 0.50 kg at a temperature of 155 degrees Celsius is dropped into an aluminum calorimeter cup of mass.200 kg containing 1.00 kg of water ( c = 4.2 X 10³ J/kg K) at 20 degrees Celsius. The system is insulated and attains equilibrium at a final temperature T_f

(I) Using the definition of specific heat capacity, set up the Q_total = 0 equation.

(II) solve for T_f

Calculate the temperature of a fluid when both Fahrenheit and a Celsius thermometer are immersed in it, under the following conditions (a) the numerical reading is identical in both thermometers, (b) the Fahrenheit is numerically twice that of the Celsius reading. Express the values in ⁰R and ⁰K.

Answer 1

Solution.

$c_a = 9.1 \sdot 10^2 J/(kg K);$

$m_a=0.50kg;$

$t_a=155^oC;$

$m_c=0.200kg;$

$m_w=1.00kg;$

$c_w=4.2\sdot10^3J/(kgK);$

$t_w=20^oC;$

$T_f-?;$

$Q_a=c_am_a(T_a-T_f);$

$Q_c=c_cm_c(T_f-T_c);$

$Q_w=c_wm_w(T_f-T_w);$

$a) If Q_{total}=0, c_am_a(T_a-T_f)+c_cm_c(T_f-T_c)+c_wm_w(T_f-T_w)=0;$

$b)c_am_a(T_a-T_f)=c_cm_c(T_f-T_c)+c_wm_w(T_f-T_w)\implies$

$T_f=\dfrac{c_a m_a T_a+c_cm_cT_c+c_wm_wT_w}{c_am_a+c_cm_c+c_wm_w};$

$T_f=\dfrac{9.1\sdot10^2\sdot0.50\sdot155+4.2\sdot10^3\sdot1.00\sdot20+9.1\sdot10^2\sdot0.200\sdot20}{9.1\sdot10^2\sdot0.50+4.2\sdot10^3\sdot1.00+9.1\sdot10^2\sdot0.200}=33^oC;$

The Fahrenheit scale and the Celsius scale "intersect" at −40 ° (−40 ° F = −40 ° C).

$-40^oC=233^oK=419.67ºR;$

Fahrenheit (° F) = Celsius (° C) * 1,8 + 32 ;

$160^oC=320^oF=433^oK=779.67ºR;$

Answer: $T_f=33^oC;$

$-40^oC=233^oK=419.67ºR;$

$160^oC=320^oF=433^oK=779.67ºR;$