Answer:m 1= 36 g=0.036 kg - the 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 λ = 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 1 1 ( 0 − 1 ) − ℎ 0 1 − 1 2 ( − 0 ) − ℎ 2 2 ( − 2 ) − 3 3 ( − 3 ) − m 1 C 1 (0−T 1 )−ice heating to 0 o C m1λ−ice melting m 1 C 2 (T−0)−melted water heating m 2 C 2 (T−T 2 )−water cooling m 3 C 3 (T−T 3 )−copper cooling 1 1 ( 0 − 1 ) + 1 + 1 2 ( − 0 ) + 2 2 ( − 2 ) + 3 3 ( − 3 ) = 0 m 1 C 1 (0−T 1 )+m1λ+m 1 C 2 (T−0)+m 2 C 2 (T−T 2 )+m 3 C 3 (T−T 3 )=0 0.036 ⋅ 2090 ⋅ 77 + 0.036 ⋅ 3.33 ⋅ 1 0 5 + 0.036 ⋅ 4186 ⋅ + 0.589 ⋅ 2090 ⋅ ( − 26 ) + 0.074 ⋅ 387 ⋅ ( − 26 ) = 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 ⋅ = 14969.368 = 10.6 1 1410.344⋅T=14969.368 T=10.61 o C Answer: = 10.6 1 T=10.61 o C
Step-by-step explanation: