o0 Sprint LTE 11:17 AM LAB-11 Heat of Fusion of Ice F... MATERIALS LabQuest Styr

Question

o0 Sprint LTE 11:17 AM LAB-11 Heat of Fusion of Ice F... MATERIALS LabQuest Styrofoam cup LabQuest App ring stand Te Probe utility clamp 600 mL beaker ice 150 mL beaker stirring rod tongs arm water 14 LAB 11 PRE LAB WORKSHEET Specific Heat Capacity (C) The energy transferred as heat that is required to raise the temperature of 1 gram of a substance by 1 kelvin. q= heat lost or gained, m = mass substance (grams) C = the Specific Heat Capapcity of a compound ( Jig x °C) 1. How much heat in Joules (J) is required to warm 2.5 L of water from 35.0 °C to 80.0°C? Assume a density of 1.00 g/mL for the water. The specific heat of water (liquid) is 4.18 J(gx C). Courses Calendar To Do Notifications Messages

Explanation / Answer

Using, q = m c dT       -- equation 1   

Where, q= heat change                                                                  , m= mass in gram,

c= specific heat (in terms of J/g0C) = 4.18 J/ g0C for water                             

dT = Final temperature T – Initial temperature

part 1: Water

q = 2500 g x (4.18 J/g0C) x (80.0 – 35.0)0C = 2500 g x (4.18 J/g0C) x 450C

q = 470250 J                ;[1 L = 1 kg water, assuming density of 1g/ mL; 1 kg = 1000 g]

part 2: Ethanol

q = 75.4 g x (2.44 J/g0C) x (58.4 – 25.1)0C = 75.4 g x (2.44 J/g0C) x 33.30C

q = 6126.4 J

part 3: Water

500 J = 40 g x (4.18 J/g0C) x (dT) = 167.2 J/0C x dT

Or, dT = 500 J / 167.2 J/0C = 2.99 0C

Thus, water temperature increased by 2.990C or 3.00C (approx.)

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