# 3.37 part B, part C and part D please help step by step. please If 3.35 the wo

Question

# 3.37 part B, part C and part D please help step by step. please

If 3.35 the would reach the highest temperature (see Table same initial of of Ais aluminum. iron, and 3.11 the copper 3.36 Substances A and B are the same mass and at you temperature. When they are heated, the final this 55 °C than the temperature of B. What does about the specific heats of I 37 the heat equation to calculate the energy for each of the equakwa 3A0 heal each th following (see 3.11) from °C 36 C oc for gof 15 a. to heat 8.5 g water from 86 to 61 calories, 5.25 18 c a. lost when 150 heat b. 2600 lost when 25 g water 15 oct 17 oC b, lost when of from to C c. 9.3 kilocalories to heat 150 of water c to 188 d. ules to heat 175 of copper from 28 the g energy for each of d. kilojo calculate the the heat equation to 3.38 Table 3.11): following (see of State from state ano ges one to 3.7 Char state when is converted it s change of a

Explanation / Answer

Answer (3.37)

The equation you use is q=mc(delta T)
q= energy; m = mass in grams; c = specific heat constant; and delta T equals change in temperature
We are solving for q (energy) basically...

Part -A.
q = (8.5)(4.18)(36 - 15)
q = 746.13 joules
Now conver joules to calories...
1 calorie = 4.184 joules
746.13 / 4.1844 =
178.31 calories

Part-B

(4.184 J/g·°C) x (25 g) x (86 - 61)°C = 2.60 x 10^3 J = 2600 J

Part-C

q = (150)(62)(4.18)
q = 38874 J
38874 J / 4.1844 =
9290.22 calories
divide by 1000 to obtain kilocalories
9.29 kilocalories

Part-D

(175g)(160)(4.18)
q = 117,040 J
KiloJoules = 117,040 / 100 =
117.04

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