Answer:
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Explanation:
Mass of ethylene glycol = m = 100 g
Specific heat capacity of ethylene glycol = c = 3.5 J/g°C
Change in temperature of ethylene glycol = ΔT
Heat loss by the ethylene glycol = Q = 350 J
[tex]Q=mc\Delta T[/tex]
[tex]\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}[/tex]
ΔT = 1°C
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
