00:01
For this problem on the topic of heat, we are told that steam at 100 degrees celsius is added to ice at 0 degrees celsius.
00:07
We want to find the amount of ice that melted and the final temperature when the mass of steam is 10 grams and the mass of ice is 50 grams.
00:15
We then want to repeat this calculation for a mass of steam of 1 gram and the mass of ice of 50 grams.
00:23
Now, firstly, we'll call q1 the heat to melt all the ice.
00:27
And this is a mass of 50 times 10 to the minus 3 kg times the latent heat 3 .33 times 10 to the power 5 joules per kg.
00:49
And so this value q1 is 1 .67 times 10 to the power 4 joules.
00:58
We'll call q2 the temperature to raise, or the heat rather to raise the temperature of ice to 100 degrees celsius, which is the mass of 50 times 10 to the minus 3 kg times the specific heat for water, which is 4 ,186 joules per kg degrees celsius times temperature change of 100 degrees celsius times temperature change of 100 degrees celsius.
01:29
Which is 2 .09 times 10 to the power of 4 joules.
01:37
Now, the total heat to melt ice and raise the temperature to 100 degrees celsius, therefore, is 3 .76 times 10 to the 4 joules...