A 21 g block of ice is cooled to -82 degrees C. It is added to 546 g of water in

Question

A 21 g block of ice is cooled to -82 degrees C. It
is added to 546 g of water in an 83 g copper
calorimeter at a temperature of 29 degrees C.
Find the final temperature. The specific
heat of copper is 387 J/kg ·
degrees C and of ice is 2090 J/kg · degrees C.
The latent heat of fusion of water is 3.33 × 105 J/kg
and its specific heatis 4186 J/kg ·degrees C.
Answer is 23.5372 degrees C. Just need to know how to do it.

Explanation / Answer

I solved a simlar question this may help .........................................................

A 40 g block of ice is cooled to -75°C and is then added to 650 g of water in an 80 g copper calorimeter at a temperature of 24°C. Determine the final temperature (in  degree C) of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C.

............................................................................................... mass of the ice block mice =40 g =40*10^-3 kg mass of water mw =650 g =650*10^-3 kg mass of copper mcu =80 g =80*10^-3 kg temperature of ice Tice =-75 0C specific heat of ice Cice =0.5 cal/g .oC =2090 J/kg .oC specific heat of water Cw =4186 J/kg .oC specific heat of copper Ccu =387 J/kg .oC process 1: warm ice to 0 0C, heat energy Q1 = miceCice(00C -Tice) ....... (1) substitute the given data in eq (1), we get Q1 =6.27*10^3 J ........................................................ process2: at 0 0C (ice is melt) heat energy Q2 =miceLf ....... (2) where, latent heat Lf =3.33*10^5 J/kg substitute the given data in eq (2), we get Q2 =1.33*10^4 J ..................................................... process 3: the block of ice to reach 0C and completely melt before the water and copper has cooled to 0C (Tf > 0C ). heat energy Q3 = [mwCw+mcuCcu][24 0C -0 0C ] ..... (3) substitute the given data in eq (3), we get Q3 =6.88*10^4 J therefore, Q3 >(Q1+Q2) at equlibrium condition, gain of heat energy equals to loss of the heat energy, i.e., Q gain =Qloss Q1+Q2+mice Cw(Tf-0 0C ) =[mwCw+mcuCcu][24 0C -Tf ] ...... (4) substitute the given data in eq (4), we get final temperature Tf = 15.92 0C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.