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0 g ice at 0^@C is converted into steam ...

`0 g` ice at `0^@C` is converted into steam at `100^@C`. Find total heat required . `(L_f = 80 cal//g, S_w = 1cal//g-^@C, l_v = 540 cal//g)`

Text Solution

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The correct Answer is:
A, B, C
`Q = mL_f+ms_w Deltatheta + mL_f`
`= m(L_f + s_wDeltatheta + L_f)`
`= 10[80 + 1 xx 100 + 540]`
`=7200 cal`
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6 gm of steam at 100^@ C is mixed with 6 gm of ice at 0^@ C . Find the mass of steam left uncondensed. (L_(f)=80 cal//g,L_(v)= 540cal//g, S_(water) = 1cal//g-^(@) C) .

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Knowledge Check

  • 1 gram of ice at -10^@ C is converted to steam at 100^@ C the amount of heat required is (S_(ice) = 0.5 cal//g -^(@) C) (L_(v) = 536 cal//g & L_(f) = 80 cal//g,) .

    A
    861 cal
    B
    12005 cal
    C
    721 cal
    D
    455 cal

  • 1 gm of ice at 0^@C is converted to steam at 100^@C the amount of heat required will be (L_("steam") = 536 cal//g) .

    A
    756 cal
    B
    12000 cal
    C
    716 cal
    D
    450 cal

  • If there is no heat loss, the heat released by the condensation of x gram of steam at 100^@C into water at 100^@C can be used to convert y gram of ice at 0^@C into water at 100^@C . Then the ratio of y:x is nearly [Given L_l = 80 cal//gm and L_v= 540 cal//gm ]

    A
    `1:1`
    B
    `2:1`
    C
    `3:1`
    D
    `2.5:1`

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    50 g of ice at 0^(@)C is converted to steam at 100^(@)C . Calculate the amount of heat given in this process. (Latent heat of vapourisation= 540 cal. per gm)

    How much work in joule is done in producing heat necessary to convert 10g of ice at -5^(@)C into steam at 100^(@)C ? Given specificheat of ice =0.5 cal g^(-1).^(@)C^(-1) , latent heat of steam = 540 cal g^(-1) .

    5 g of steam at 100^(@)C is mixed with 10 g of ice at 0^(@)C . Choose correct alternative /s) :- (Given s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g )

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