100g of cu block at 100°c is dropped in a calorimeter containing 180g of oil at 25°c .If 10% of the total heat of cu block is to the surrounding . What is the final temperature of the mixture (Assuming the S.H.C of oil and cu are 2jg-¹k-¹ and 0.4jg-¹k-¹)?

Robert01

4 years ago

The amount of heat lost is 90% efficient to the system.

Mcu= 100g
T1= 100°C
Mo= 180g
T2= 25°C
T3= ?
co= 2J/g-k
ccu= 0.4J/g-k

Heat lost = Heat gained

90% × Mcu × ccu × (T1 – T3) = (Mcal. × ccal. + Mo × co)(T3 – T2)

0.9 × 100 × 0.4 × (100 - T3)= (180 × 0.4 + 180 × 2)(T3 - 25)

36(100 - T3)= 432(T3 - 25)

3600 + 10800= 432T3 + 36T3

14400= 468T3

T3= 30.77°C.