Question

10. An iron pan is used to fry a cylindrical piece of meat (mass=0.120kg, 1.50cm thickness and 10.0cm diameter). The temperature of the pan is heated to a constant temperature of 350.0°F. The thermal conductivity of the meat is 0.400W/(mK) and the emissivity of the meat is 0.900. Assume the specific heat capacity of the meat is 3500J/(kg:K). a. After the meat is placed on the pan, the temperature of the contacting surface is raised to a constant temperature Troom=70.0°F of 100.0°C and the opposite face of the meat initially is 10.0cm 5.0°C. Calculate this initial rate of conductive heat flow through the meat. Assume steady-state heat transfer. 1.50cm meat Hot Pan b. When enough heat has passed through the meat, the opposite face of the meat will reach the temperature of the room (70.0°F). What is the temperature at the center of the meat? c. Determine the rate of conductive heat flow through the meat in (b)? d. How much (net) heat has the meat absorbed by the time its upper surface reaches room temperature?

Answer 1

at 52 4 1.5x10-2 =99.14 Jls Arx @ The rate of heat conduction is given by the formula del 2-KAOTK= thermal Conductivity=0.400 Ac area ferodo section TD2 Tz a temp.diffrence=1005 oaz change in length. 1.50 =95 Rate of heat flow thoroughe heat from pan to upper surface -0.400 X TT XO-12x 95 dt 4 19.8.96 ~ 198.95 Is And © Convert 70°F to °C ») (70-32153 211CA By lin earraniation! ap it to steady state heat tomater Temp. gradient = 100-21.1 1.800 Now temp. of Center = 78.960.75 1.5 Now temp.of surface ip 21.1°C ; OT=7898 How anductive rate of heat transfert do It 21.1° low'c = 78-9 = 39.450 2 Ano 02

@ Initially tenk of the treat in considermigat 5c I am considering that initially the whole wap at 5c. Now by linear variation temp as a function 1.5,210) (Tamo man ko) = 100-21.1(2-0) of a =) 21.1°C 001:5. Co 110 ° T= 100 &-526(a-6 = 100 – 52.6 x 789 T 100-52-62 Now Consider a differential didc of thick neopdr. CHA A - TTM, len after cuttal valumed V-nex dr = TTD 2 dxi it maso &m=0.120kg density St m = ma ✓ TIDAL Now mesoof differential dive=dm-8 dv dm = 4m 2 dando 84 Now heat absorbed by this digca) d8=0ms ft-Ti) = m. dets(160–9164-5) me cassda-s2-cfed x) oge 3-120.935pour (95815X102–240 x65x1333) 1.58102

2 69=0:120X360vw (asuis x16 (astisxio² – 526 XUS 103 1-5X162 X (1 1425 – 59.175*162 162 ) 3800 = 0.120 1-5 = 2 80x (142.5-0.89 =39898345 Ang