Question

please answer the question. thank you.

Trial Trial 2 mass of-tre dreel 43.499 3.৭৭ mass atwacciaualaiener 30.09 90.0c 30.03 ompecaus.at net reta demwerant.o.tCOO Maicc au.1c au C tinal emp Caiec.ondreta) a3.4 c 30.1C Tempecantt cnaneLaT Dnial Temyeradae Chonge AT) at waier -2.6C -৬০. o.9c 3.2 C Specielcroct ot metal in Trial=0831°c Specitic reatot retal inTrial 2 o.35 ° Prevage speiticreat o.a1 l9e Metal Rb Density e3glmL cavses a 140.0 mL OVESTION . 1t a sanple of your medalat 190.0 sample ot uaderto inucAse in lemperechurt fron 23.0c to a4.30unatisthe mais of he retal? ve the aaage specki heat

Answer 1

Average Specific heat of the metal as calculated = $C_{metal} = 0.21 \ J \cdot g^{-1} \cdot ^\circ C^{-1}$

Note that actual value might differ as the values in trial 1 and trial 2 are quite different. But we will do the calculation with the above value.

Starting temperature of the metal = $T_{i,metal} = 190^\circ C$

Starting temperature of water = $T_{i,water} = 23.0 ^\circ C$

Final temperature of metal and water system = $T_{f} = 24.3 ^\circ C$

Assuming that the heat lost by hot metal is completely gained by the water without any loss to the surrounding, we can write the following equation for energy balance

$q_{metal, lost} = q_{water, gain} \\ \Rightarrow m_{metal} \times C_{metal} \times \left | T_f - T_{i,metal} \right | = m_{water} \times C_{water} \times \left | T_f - T_{i,water} \right |$

Note that the specific heat of water is $C_{water}= 4.184 \ J \cdot g^{-1} \cdot ^\circ C^{-1}$

Density of water = 1 g/mL.

Hence, mass of 140.0 mL of water is $m_{water} = 140.0 \ g$

Now, putting in values in the above equation,

$q_{metal, lost} = q_{water, gain} \\ \Rightarrow m_{metal} \times C_{metal} \times \left | T_f - T_{i,metal} \right | = m_{water} \times C_{water} \times \left | T_f - T_{i,water} \right | \\ \Rightarrow m_{metal} \times 0.21 \ J \cdot g^{-1} \cdot ^\circ C^{-1} \times \left | 24.3 ^\circ C - 190 ^\circ C \right | = 140.0 \ g \times 4.184 \ J \ g^{-1} \ ^\circ C ^{-1} \times \left | 24.3 ^\circ C - 23.0 ^\circ C \right | \\ \Rightarrow m_{metal} = \frac{761.488 \ J}{ 0.21 \ J \cdot g^{-1} \cdot ^\circ C^{-1} \times \left | 24.3 ^\circ C - 190 ^\circ C \right |} \\ \\ \Rightarrow m_{metal} \approx 21.88 \ g$

Hence, the mass of metal is about 21.88 g .