1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The specific heat of water is $4200\mathrm{J}{\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$ and the latent heat of ice is $3.4×105\mathrm{J}{\mathrm{kg}}^{-1}$.$100\mathrm{gm}$ of ice at $0°\mathrm{C}$ is placed in $200\mathrm{g}$ of water at $25°\mathrm{C}$. The amount of ice that will melt as the temperature of water reaches $0°\mathrm{C}$ is close to (in grams):

A

$63.8$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

$64.6$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

$61.7$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

$69.3$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C $61.7$The explanation for the correct option:C: $61.7$Heat loss by water when it cools down to ${0}^{0}\mathrm{C}$ is,The formula used for the calculation:$\begin{array}{rcl}\mathrm{Q}& =& \mathrm{mws}∆\mathrm{\theta }\\ & =& \left(200/1000\right)×\left(4200\right)×\left(25\right)\\ & =& 21000\mathrm{J}\end{array}$Which is less than the amount of heat $\left({\mathrm{m}}_{\mathrm{i}}\mathrm{f}\right)$ required to melt ice completely.$∆{\mathrm{m}}_{\mathrm{i}}\mathrm{L}=21000\mathrm{J}$To find the amount of ice melt (∆mi), take:$\begin{array}{rcl}∆{\mathrm{m}}_{\mathrm{i}}& =& \left[21000/\left(3.4×105\right)\right]\mathrm{x}103\mathrm{gm}\\ & =& 61.7\mathrm{grams}\end{array}$Therefore, option (C) $61.7$ is the correct option.

Suggest Corrections
30
Join BYJU'S Learning Program
Related Videos
Specific heat
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program