Question

calculate the specific heat of metal piece. if 500gm metal piece at 300℃ is dropped in 5kg of water. its temperature raises from 30℃ to 75℃(there is no heat loss)​

Answer 1

Answer:

For metal,

mass of metal (m1) = 500 gm = 0.5 kg

temperature of metal (T1) = 300°C

final temperature (T) = 75°C

For water,

mass of water (m2) = 5 kg

specific heat of water (s2) = 4200

initial temperature (T2) = 30°C

final temperature (T) = 75°C

•QUESTION•

specific heat ( s)= ?

•ANSWER•

we know,

according to the principles of calorimetry,

$q _{1} = q _{2} \\ = > m _{1}s _{1} (t _{1} - t) = m _{2}s _{2}(t - t _{2}) \\ = > 0.5 \times s _{1} \times (300 - 75) = 5 \times 4200 \times (75 - 30) \\ = > 0.5 \times s _{1} \times225 = 945000 \\ = >112.5 \times s _{1} = 945000 \\ = > s _{1} = 8400$

ANS: 8400