Question

A machine gear consists of 0.10 kg of iron and 0.16 kg of copper.

How much total heat is generated in the part if its temperature increases by 35 C°? (Specific heats of iron and copper are 450 and 390 J/kg⋅°C, respectively.)
a. 910 Jb. 3 800 Jc. 4 000 Jd. 4 400 J

Answer 1

option (c)

Step-by-step explanation:

mass of iron = 0.10 kg

mass of copper = 0.16 kg

rise in temperature, ΔT = 35°C

specific heat of iron = 450 J/kg°C

specific heat of copper = 390 J/kg°C

Heat by iron (H1) = mass of iron x specific heat of iron x ΔT

H1 = 0.10 x 450 x 35 = 1575 J

Heat by copper (H2) = mass of copper x specific heat of copper x ΔT

H1 = 0.16 x 390 x 35 = 2184 J

Total heat H = H1 + H2

H = 1575 + 2184 = 3759 J

by rounding off

H = 4000 J

Answer 2

Answer:b

Step-by-step explanation:

Given

mass of iron Gear
`m_(iron)=0.1 kg`

mass of copper
`m_(cu)=0.16 kg`

specific heat of iron
`c_(iron)=450 J/kg-^(\circ)C`

specific heat of copper
`c_(cu)=390 J/kg-^(\circ)C`

heat gain by iron gear
`=m_(iron)c_(iron)\Delta T`

`Q_1=0.1* 450* 35=1575 J`

heat gain by iron gear
`=m_(cu)c_(cu)\Delta T`

`Q_2=0.16* 390* 35=2184 J`

Total heat
`Q_(net)=Q_1+Q_2`

`Q_(net)=1575+2184=3759 J\approx 3800 J`