Question

steam is passed to the 40 gram of the ice at -10 degree Celsius the temperature of the mixture become 80 degree celsius is also mass of mixture will be approximately​

Answer 1

Let the two bodies have masses m1, m2 and specific heats s1 and s2. Then

Let the two bodies have masses m1, m2 and specific heats s1 and s2. Thenm1s1=m2s2  or m2m1=s1s2

Let the two bodies have masses m1, m2 and specific heats s1 and s2. Thenm1s1=m2s2  or m2m1=s1s2Let s= specific heat of the composite body.

Let the two bodies have masses m1, m2 and specific heats s1 and s2. Thenm1s1=m2s2  or m2m1=s1s2Let s= specific heat of the composite body.Then (m1+m2)s=m1s1+m2s2=2m1s1

Let the two bodies have masses m1, m2 and specific heats s1 and s2. Thenm1s1=m2s2  or m2m1=s1s2Let s= specific heat of the composite body.Then (m1+m2)s=m1s1+m2s2=2m1s1s=m1+m22m1s1=m1+m1(s2s1)

hope it helps uu mate

Answer 2

Answer:

52

Explanation:

m1Lf+m1sdT=m2Lv

40×80+40×1×(80-0)=m2×540

6400/540=m2

m2=12

m of mixture=40+12=52