I need help with these 5 questions.

SOLUTION Now we'll calculate the approximate amount of heat energy needed to raise body temperature a given SET UP AND SOLVE The temperature change is AT = 39.0' C - 37.0.C = 2.0 Co = 2.0 K. The amount. During a bout with the flu, an 80 kg man specific heat of water is 4190 J/ (kg . K) . From the equation Q = mcAT, ran a fever of 2.0 Co ; that is, his body temperature was 39.0' C (or 102.2 F) instead of the normal Q = mcAT = (80 kg) [4190 J/ (kg . K)](2.0 K) 37.0' C (or 98.6. F). Assuming that the human = 6.7 x 105 J body is mostly water, how much heat was required to raise his temperature by that amount? REFLECT This amount of heat is equal to 1.6 x 10 cal, or 160 kcal (160 food calories). This is roughly half the calorie intake from eating 1 ounce of milk chocolate. Part A - Practice Problem: If a 31 kg child has a temperature of 104 F = 40 C, how much heat was needed to raise her temperature to that value from an initial value of 37 C? Express your answer in joules. VO AEd Submit Previous Answers Request AnswerV Part A An insulated beaker with negligible mass contains liquid water with a mass of 0.280 kg and a temperature of 69.1 . C . How much ice at a temperature of -21.2 C must be dropped into the water so that the final temperature of the system will be 23.0 . C ? Take the specific heat of liquid water to be 4190 J/kg . K , the specific heat of ice to be 2100 J/kg . K , and the heat of fusion for water to be 3.34x105 J/kg . > View Available Hint(s) AED ? mice = kg SubmitThis example involves heat flow between a liquid and its container. A Turkish restaurant serves coffee in copper mugs (with the inside tin plated to avoid copper poisoning). Awaiter fills a cup having a mass of 0.100 kg and initially at 200C with 0.300 kg of coffee that is initially at 700C. What is the final temperature after the coffee and the cup attain thermal equilibrium? (Assume that coffee has the same specific heat as water and that there is no heat exchange with the surroundings.) 50LU110N SET UP AND SOLVE Call the nal temperature T. The (negative) change in temperature of the coffee is ATmrrea = T i 70C, and the {positive} change in temperature of the cup is ATW. 2 T i 20C. We need to find the heat lost by the coffee and gained by the copper cup and then set the algebraic sum equal to zero. Because the coffee loses heat, it has a negative value of Q: Qcoffee = mcwater ATeoffee (0.300 kg)[4190 J/(kg -K)](T 700C) (Note that ATmffee is negative in this expression.) The heat gained by the copper cup is momma, TAT\Part A Each cycle, a certain heat engine expels 270 J of heat when you put in 335 J of heat. Find the efficiency of this engine. Express your answer as a percentage. IVO AE e = Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part B Find the amount of work you get out of the 335 J heat input. Express your answer in joules. IVO A : Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Determine the gauge pressure for a 2.0-cm-diameter faucet located next to the washing machine in the basement, 2.7 m below where the pipe enters the house. Express your answer with the appropriate units. HA To second floor (1 cm pipe Value Units 5.0 m Submit Previous Answers Request Answer From water Hot-water 1 supply tank (2 cm pipe)