Question

For calculations, assume Density of HCI = 1.0 g/mL Cs of soin = 4.184 / (ec) Calorimetry PART A. Determination of the Calorimeter Constant Mass, "Hot" water 50mL 21.9°C 39.fec "Cold" water 50 mL los.ooc 20 OC Initial Temperature, Final Temperature, Change in temp, 9=mCH.047 9 = m 047 9cal = CAT. w+or+9 = 0 (show work in the box) Heat of hot water, Heat of cold water, qew Heat of calorimeter, Coal Co

Answer 1

From the given data ,

Density of water = 1g/ml

Mass/volume = 1g/ml

Mass/50ml = 1g/ml

Mass = 50ml×1g/ml

Mass = 50g

So,

Mass of water = 50g

Step (1) Calculation for haat of hot water.

m_{​​​​​​hw} = 50g

Specific heat of water

C_{​​​​​​S(H2O)} = 4.184J/g°C

Change in temperature

∆T_hw = -25.4°C

Putting the all value in given formula

q_{​​​​​hw} = m_hwC_s(H2O)∆T_hw

q_{​​​​​hw} = (50g)(4.184J/g°C)(-25.4°C)

q_{​​​​​​hw} = 5313.68J

Step(2) Calculation for Heat of cold water.

m_{​​​​​cw} = 50g

Specific heat

C_{​​​​​​S(H2O)} = 4.184J/g°C

Change in temperature

∆T_cw = 17.7°C

Putting the all value in given formula

q​​​​​​_{​​​​​​Cw} = m_cwC_s(H2O)∆T_cw

q_{​​​​​​cw} = (50g)(4.184J/g°C)(17.7°C)

q_{​​​​​​cw} = 3702.84J

Step(3) Calculation for heat of calorimeter

Putting the all value in given formula

q_{​​​​​​hw} + q_{​​​​​qw} + q_{​​​​​​cal} = 0

-5313.68J + 3702.84J + q_{​​​​​​cal} = 0

-1610.84J + q_{​​​​​​cal} = 0

q_{​​​​​​cal} = 1610.84J