Question

Please be very detailed when explaining this I want to understand to the best of my ability.

Q 18.11: The energy given off by 300 g of an alloy as it cools by 50°C raises the temperature of 300 g of water from 30°C to 40°C. The specific heat of the alloy is A 0.10 cal/g/°C B 0.20 cal/g/°C C 5.0 cal/g/°C 0.50 cal/g/°C E 0.15 cal/g/°C

Answer 1

Given,

The mass, m₁ = 300 g = 0.3 kg

m₂ = 300 g = 0.3 kg

$\Delta$T₁ = 50 ⁰C

$\Delta$T₂ = 40 - 30 = 10 ⁰C

The specific heat of the alloy is,

Q = mc$\Delta$T

So,

m₁c₁$\Delta$T₁ = m₂c₂$\Delta$T₂

300 * c₁ * 50 = 300 * 1 * 10

c₁ = 0.2 cal / g⁰C

Answer: B) 0.2 cal / g⁰C