Question

10g water at 80 degree is mixed with 50g copper at 10 degree .find temperature of mixture c water =4.2 j/g c copper=0.4 j/g

Answer 1

Answer:

57.41°C

Explanation:

Let the final temperature be t

By the principle of calorimetry,

Heat lost = heat gained

m₁c₁(t₁-t) = m₂c₂(t-t₂)

m₁ = 10g water                                                              m₂= 50g copper

c₁ =4.2 Jg⁻¹°C⁻¹                                                              c₂ = 0.4Jg⁻¹°C⁻¹

t₁ = 80°C                                                                         t₂ = 10°C

Substituting,

10 X 4.2 X (80-t)        =         50 X 0.4 X (t-10)

42 X (80-t)                 =          20 X (t-10)

3360 - 42t                 =           20t - 200

3560                           =           62t

t = 3560/62 =57.41°C