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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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