$0.1\,m^3$ of water at $80\,^oC$ is mixed with $0.3\,m^3$ of water at $60\,^oC$. The final temperature of the mixture is ........ $^oC$

A drilling machine of $10\,KW$ power is used to drill a bore in a small aluminium block of mass $8\,kg.$ If $50\%$ of power is used up in heating the machine itself or lost to the surroundings then  ........ $^oC$  is the rise in temperature of the block in $2.5\,minutes$

[specific heat of aluminium $= 0.91\,J/g\,\,^oc$ ]

Find the quantity of heat required to convert $40\; gm$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L _{\text {ice }}$ $=0.336 \times 10^6 J / kg$.

specific heat of ice $=2100 \;J / kg - K$ sp heat of water= $4200\; J / kg - K$

Three liquids with masses ${m_1},\,{m_2},\,{m_3}$ are thoroughly mixed. If their specific heats are ${c_1},\,{c_2},\,{c_3}$ and their temperatures ${T_1},\,{T_2},\,{T_3}$ respectively, then the temperature of the mixture is

A $20 \,g$ bullet whose specific heat is $5000 \,J kg ^{\circ} C$ and moving at $2000 \,m / s$ plunges into a $1.0 \,kg$ block of wax whose specific heat is $3000 \,J kg ^{\circ} C$. Both bullet and wax are at $25^{\circ} C$ and assume that $(i)$ the bullet comes to rest in the wax and $(ii)$ all its kinetic energy goes into heating the wax. Thermal temperature of the wax $\left(\right.$ in $\left.^{\circ} C \right)$ is close to

The specific heat of water $=4200\, J\, kg ^{-1}\, K ^{-1}$ and the latent heat of ice $=3.4 \times 10^{5}\, J\, kg ^{-1}.$ $100$ grams of ice at $0^{\circ} C$ is placed in $200\, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} C$ is close to (in $grams$)