Question

A mass of (200 of hot water at (75, 0 degrees * C) is mixed with cold water of mass M at (5, 0 degrees * C) The final temperature of the mixture is (25, 0 degrees * C) What is the mass of the cold water (M)

Answer 1

∆T for hot water

• 75-25
• 50°C

∆T for cold water

• 25-5
• 20°C

Both are equal according law of conservation of energy

$\\ \rm\Rrightarrow m_1c\delta T_0=m_2c\delta$

$\\ \rm\Rrightarrow 200(50)=m_2(20)$

$\\ \rm\Rrightarrow 10000=m_2(20)$

$\\ \rm\Rrightarrow m_2=500g=0.5kg$

Answer 2

Q in = Q out

mcΔt in = mcΔt out (c = equal)

mΔt in = mΔt out

200 x  (75-25) = M x (25-5)

M = 500 g