Practice portal › Thermal Properties of Matter › Calorimetry and Specific Heat
Asked in JEE Main 23rd Jan 1st Shift 2026 · Mechanical work converted into heat
Given: m₁=15 kg at +10 m s⁻¹ and m₂=25 kg at -30 m s⁻¹; c=31×4.2=130.2 J kg⁻¹ ° C⁻¹.
Momentum is conserved: v=(15(10)+25(-30))/(40)=(-600)/(40)=-15 m s⁻¹.
Kᵢ=1/2(15)(10)²+1/2(25)(30)²=750+11250=12000 J, and K_f=1/2(40)(15)²=4500 J.
Heat produced =Kᵢ-K_f=7500 J (the same as 1/2μ vᵣₑₗ²=1/2(9.375)(40)², a useful check).
Both spheres keep the heat, so Δ T=(7500)/(40×130.2)=1.44 °C.
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