InterviewSolution
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InterviewSolution
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The following graph represents change of state of `1` gram of ice at `-20^@C`. Find the net heat required to convert ice into steam at `100^@ C`. `S_(ice) = 0.53 cal//g -^@ C` . |
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Answer» In the figure : `a to b` : temperature of ice increases until its reaches its melting point `0^@ C` `Q_(1)=mS_(ice)[0-(-20)] =(1)(0.53)(20)=10.6 cal` `b to c` : Temperature remains constabt until all the ice has melted. `Q_(2) = mL_(f) = (1)(80) = 80 cal` `c to d` : Temperature of water againm rises until it reaches its boiling point `100^@C` `Q_(3) = mS_(water) [100-0] = (1)(1.0)(100) = 100 cal` `d to e` : Temperature is again constant until all the water is transformed into the vapour phase `Q_(4) = mL_(v) = (1)(530) = 539 cal` Thus, the net heat required to convert `1 g` of ice at `-20^@ C` into steam at `100^@ C` is `Q = Q_(1) + Q_(2) + Q_(3) + Q_(4) = 729.6 cal`. |
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