Question

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

Answer 1

Answer:

option (c)

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

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 $=m_{iron}c_{iron}\Delta T$

$Q_1=0.1\times 450\times 35=1575 J$

heat gain by iron gear $=m_{cu}c_{cu}\Delta T$

$Q_2=0.16\times 390\times 35=2184 J$

Total heat $Q_{net}=Q_1+Q_2$

$Q_{net}=1575+2184=3759 J\approx 3800 J$