Answer
a) It raises the temperature of the ice to $-15^\circ\rm C$.
b) It melts 3 g of the ice.
c) It raises the temperature of the liquid water to $12.4^\circ\rm C$.
d) It evaporates 0.442 g of the water.
e) It raises the steam temperature to $115^\circ\rm C$.
Work Step by Step
a)
Let's calculate the final temperature of the ice when we add an amount of heat of 1000 J to 100 g of ice at $-20^\circ \rm C$ to see if it is enough to melt it or not.
$$Q=mc_{ice}\Delta T=mc_{ice}\left(T_f-T_i\right)$$
So,
$$T_f=\dfrac{Q}{mc_{ice}}+T_i\tag 1$$
$$T_f=\dfrac{1000}{100\times 10^{-3}\cdot 2100}-20$$
$$T_f=\boxed{\color{red}{\bf-15}^\circ \rm C}$$
It raises its temperature to $-15^\circ\rm C$.
b)
In this case, when the ice is at $0^\circ \rm C$, adding heat will melt the ice.
We need to find how much of the 100-g ice will melt.
$$Q=m_{melted}L_f$$
$$m_{melted}=\dfrac{Q}{L_f}=\dfrac{1000}{333\times10^3}=3\times 10^{-3}\;\rm kg$$
$$m_{melted}=\boxed{ \color{red}{\bf3}\ \;\rm g}$$
c)
We need to find the final temperature of the water.
Using equation (1) but for liquid water;
$$T_f=\dfrac{Q}{mc_{water}}+T_i$$
$$T_f=\dfrac{1000}{0.1\cdot 4186 }+10$$
$$T_f=\boxed{\color{red}{\bf12.4}^\circ \rm C}$$
It raises the temperature of the liquid water to $12.4^\circ\rm C$.
d)
Adding heat to the water at $100^\circ \rm C$ will evaporate some of it. We need to find the amount that evaporates.
$$Q=m_{evaporated}L_v$$
$$m_{evaporated}=\dfrac{Q}{L_v}=\dfrac{1000}{2260\times 10^3}=4.42\times 10^{-4}\;\rm kg$$
$$m_{evaporated}=\boxed{\color{red}{\bf0.442}\;\rm g}$$
e)
Adding heat to steam will raise its temperature.
Using equation (1) but for steam;
$$T_f=\dfrac{Q}{mc_{steam}}+T_i $$
$$T_f=\dfrac{1000}{0.1\cdot 2010}+110$$
$$T_f=\boxed{\color{red}{\bf115}^\circ \rm C}$$
It raises the steam temperature to $115^\circ\rm C$.