Answer to Question #107371 in Physics for Nyx

Question

A 36 g block of ice is cooled to −77 ◦C. It
is added to 589 g of water in an 74 g copper
calorimeter at a temperature of 26◦C.
Find the final temperature. The specific
heat of copper is 387 J/kg ·
◦C and of ice is
2090 J/kg · ◦C . The latent heat of fusion of
water is 3.33 × 105
J/kg and its specific heat
is 4186 J/kg · ◦C .
Answer in units of ◦C.

Answer 1

m1= 36 g=0.036 kg - mass of ice

T1= −77 ◦C. - temperature of ice

m2=589 g=0.589 kg - mass of water

C1=2090 J/kg · ◦C - specific heat of ice

$\lambda$ = 3.33 × 10^5 J/kg - latent heat of fusion of water

T2=26◦C. - temperature of water

C2= 4186 J/kg · ◦C . - - specific heat of water

m3=74 g=0.074kg - mass of copper

T3=26◦C. - temperature of copper

C3=387 J/kg ·◦C - specific heat of copper is and of ice is

T - ? - final temperature

$m_1C_1(0-T_1) - ice\ heating\ to \ 0^oC\\ m1\lambda - ice\ melting\\ m_1C_2(T-0) - melted\ water\ heating\ \\ m_2C_2(T-T_2) - water\ cooling\\ m_3C_3(T-T_3) - copper\ cooling$

$m_1C_1(0-T_1)+m1\lambda+m_1C_2(T-0)+m_2C_2(T-T_2)+m_3C_3(T-T_3)=0\\$

$0.036 · 2090 · 77+0.036 · 3.33 · 10^5+0.036 · 4186 · T+0.589 · 2090· (T-26)+0.074·387· (T-26) =0$

$1410.344 · T=14969.368\\ T=10.61^oC$

Answer: $T=10.61^oC$