10g of ice -20-o-C is added to 10g of water -50-o-C- Specific heat of water -1cal-g -o-C- specific heat of ice-0-5cal-g-o-C- Latent heat of ice -80cal-g- Then resulting temperature is -20-o-C-15-o-C-0-o-C-50-o-C

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Toppr Admin · Accepted Answer

10g of ice at −20oC is added to 10g of water at 50oC. Specific heat of water =1cal/goC, specific heat of ice=0.5cal/goC. Latent heat of ice =80cal/g. Then resulting temperature is -−20oC15oC0oC50oC

$10 g$ of ice at $- 20^{o} C$ is added to $10 g$ of water at $50^{o} C$ . Specific heat of water $= 1 c a l / g^{o} C$ , specific heat of ice $= 0.5 c a l / g^{o} C$ . Latent heat of ice $= 80 c a l / g$ . Then resulting temperature is -
$- 20^{o} C$
$15^{o} C$
$0^{o} C$
$50^{o} C$