10 gm of ice at – 100 C is mixed with 100 gm of water at 500 C contained in a calorimeter weighing 50gm. (Specific heat of water = 1 cal. gm–1. 0C–1, Latent heat of ice=80cal/gm, specific heat of ice = 0.5 cal/gm / 0C and specific heat of copper = 0.09 cal/gm/0C). The final temperature reached by the mixtures is

Heat gained by ice= Heat lost by water + Heat lost by calorimetermsdT+mL+msdTice=(msdT)w+(msdT)c[(10)(0.5(0−(−10)]+(10)(80)+(10)(1)(T−0) =(100)(1)(50−T)+50(0.09)(50−T)T=38.2∘C

10 gm of ice at – 100 C is mixed with 100 gm of water at 500 C contained in a calorimeter weighing 50gm. (Specific heat of water = 1 cal. gm–1. 0C–1, Latent heat of ice=80cal/gm, specific heat of ice = 0.5 cal/gm / 0C and specific heat of copper = 0.09 cal/gm/0C). The final temperature reached by the mixtures is

Heat gained by ice= Heat lost by water + Heat lost by calorimetermsdT+mL+msdTice=(msdT)w+(msdT)c[(10)(0.5(0−(−10)]+(10)(80)+(10)(1)(T−0) =(100)(1)(50−T)+50(0.09)(50−T)T=38.2∘C

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

10 gm of ice at – 10⁰ C is mixed with 100 gm of water at 50⁰ C contained in a calorimeter weighing 50gm. (Specific heat of water = 1 cal. gm^–1. ⁰C^–1, Latent heat of ice=80cal/gm, specific heat of ice = 0.5 cal/gm / ⁰C and specific heat of copper = 0.09 cal/gm/⁰C). The final temperature reached by the mixtures is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

38.2⁰ C

40.0⁰ C

25.5⁰C

30.0 ⁰C

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Heat gained by ice
= Heat lost by water + Heat lost by calorimeter
$\begin{array}{l} (msdT + mL + {msdT}_{ice} = (msdT)_{w} + (msdT)_{c} \\ [(10) (0 .5 (0 - (- 10)] + (10) (80) + (10) (1) (T - 0) \\ = (100) (1) (50 - T) + 50 (0 .09) (50 - T) \\ T = 38 {.2}^{\circ} C \end{array}$