I need help with 13 and 14. Quantity of Heat and Calorimetry. If 0.30 kg of tea

Question

I need help with 13 and 14.

Quantity of Heat and Calorimetry. If 0.30 kg of tea at 203 degree C is poured into a 150 grams glass cup initially at 25 degree C, what will be the common final temperature T_f(in degree C) of the tea and cup when equilibrium is reached, assuming no heat flows to the surroundings? 86 deg c 89 deg C 87 deg C 088 deg C Pascal's Principle. Suppose the hydraulic has a small cylindrical piston with diameter 11 cm and a larger piston with radius 21 cm. The mass of a car placed on the larger piston is 1020 kg. Assuming the two pistons are at the same height what force must be applied to the small piston to lift the car? 1.5 times 10^5 N 7.0 times 10^1 N 3.6 times 10^4 N 6.9 times 10^2 N

Explanation / Answer

Question 13

given

mass of tea m1 = 0.30 kg, tempereature T1 = 203 0C

mass of glass cup is m2 = 0.150 kg, at temperature T2 = 25 0C

specific heat capacities of tea and water C1 =4186 J/kg 0C

of cup is C2 = 840 j/kg 0C

here the tea will loose heat energy and cup gain heat energy

and there is no heat flow to the surroundings so calorimetri principle says that

heat energy lost by tea is = heat energy gain by cup

we know that Q = mc (T2-T1)

Q1 = m1*c1(T1-T) and Q2 = m2*C2(T-T2)

here Q1 = Q2

m1*c1(T1-T) = m2*C2(T-T2)

substituting the values

0.3*4186(203-T) = 0.150*840(T-25)

T = 169.91 0C = 170 0C

Question 14.

Given mass of the car m = 1020 kg

weight (force) is W = mg = 1020*9.8 N

raidus of small cylindrical piston is r1 = 0.055 m ,

raidus of large cylindrical piston is r2 = 0.21 m ,

pascal's law states that the pressure is same throught the fluid that is

F1/A1 = F2/A2

F1 =F2*A1/A2

F1 = 1020*9.8*(pi*0.055^2)/(pi*0.21^2) N

F1 = 685.666 N

F1 = 6.9*10^2 N

answer is Last option <<<<<-----------------------

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